Problem #1
A student drops a ball from a window 2.0 m above the sidewalk. The ball accelerates at 9.80 m/s2. How fast is it moving when it hits the sidewalk?
Answer:
Given: initial velocity vi = 0, initial position yi = 0, final position yf = 2.0 m,
Find: final velocity, vf ?
we use formulas to determine the magnitude of position
yf = yi + vit - ½gt2
2.0 m = 0 + 0 x t – ½(9,8 m/s2)t2
9.8t2 = 2.0
t2 = 1/(4.9)
t = 1/0.7 = 10/7 detik
we use formulas to determine the magnitude of final velocity
vf = vi – gt = 0 – 9.8 m/s2 x 10/7 s
vf = -14 m/s
Problem #2
(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 50 m? (b) How long will it be in the air?
Answer:
(a) First, we decide on a coordinate system. I will use the one shown in Figure, where the y axis points upward and the origin is at ground level. The ball starts its flight from ground level so its initial position is yi = 0. When the ball is at maximum height its coordinate is yf = 50 m, but we also know its velocity at this point. At maximum height the instantaneous velocity of the ball is zero. So if our “final” point is the time of maximum height, then vf = 0. So for the trip from ground level to maximum height, we know yi, yf, vf and the acceleration g = -9.8 m/s2 but we don’t know vi or the time t to get to maximum height.
we use formulas to determine the magnitude of final initial velocity
vf2 = vi2 - 2g(yf – yi)
02 = vi2 – 2(9.8 m/s2)(50 m – 0)
vi2 = 980 m2/s2
The next step is to “take the square root”. Since we know that vi must be a positive number, we know that we should take the positive square root of 980 m2/s2. We get:
vi = +31 m/s
The initial speed of the ball is 31 m/s.
(b) We want to find the total time that the ball is in flight. What do we know about the ball when it returns to earth and hits the ground? We know that its y coordinate is equal to zero. (So far, we don’t know anything about the ball’s velocity at the the time it returns to ground level.) If we consider the time between throwing and impact, then we do know yi, yf, vi and of course a. We use formulas to determine the magnitude of final initial time,
yf = yi + vit - ½gt2
0 = 0 + (31 m/s)t – ½ (9.8 m/s2)t2
It is not hard to solve this equation for t. We can factor it to give:
t [31 – 4.9t) = 0
t = 31/4.9 = 6.4 s
The ball spends a total of 6.4 seconds in flight.
Problem #3
A ball is thrown directly downward with an initial speed of 8.00 m s from a height of 30.0 m. When does the ball strike the ground?
Given: initial velocity vi = 0, initial position yi = 0, final position yf = 2.0 m,
Find: final velocity, vf ?
we use formulas to determine the magnitude of position
yf = yi + vit - ½gt2
2.0 m = 0 + 0 x t – ½(9,8 m/s2)t2
9.8t2 = 2.0
t2 = 1/(4.9)
t = 1/0.7 = 10/7 detik
we use formulas to determine the magnitude of final velocity
vf = vi – gt = 0 – 9.8 m/s2 x 10/7 s
vf = -14 m/s
Problem #2
(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 50 m? (b) How long will it be in the air?
Answer:
(a) First, we decide on a coordinate system. I will use the one shown in Figure, where the y axis points upward and the origin is at ground level. The ball starts its flight from ground level so its initial position is yi = 0. When the ball is at maximum height its coordinate is yf = 50 m, but we also know its velocity at this point. At maximum height the instantaneous velocity of the ball is zero. So if our “final” point is the time of maximum height, then vf = 0. So for the trip from ground level to maximum height, we know yi, yf, vf and the acceleration g = -9.8 m/s2 but we don’t know vi or the time t to get to maximum height.
we use formulas to determine the magnitude of final initial velocity
vf2 = vi2 - 2g(yf – yi)
02 = vi2 – 2(9.8 m/s2)(50 m – 0)
vi2 = 980 m2/s2
The next step is to “take the square root”. Since we know that vi must be a positive number, we know that we should take the positive square root of 980 m2/s2. We get:
vi = +31 m/s
The initial speed of the ball is 31 m/s.
(b) We want to find the total time that the ball is in flight. What do we know about the ball when it returns to earth and hits the ground? We know that its y coordinate is equal to zero. (So far, we don’t know anything about the ball’s velocity at the the time it returns to ground level.) If we consider the time between throwing and impact, then we do know yi, yf, vi and of course a. We use formulas to determine the magnitude of final initial time,
yf = yi + vit - ½gt2
0 = 0 + (31 m/s)t – ½ (9.8 m/s2)t2
It is not hard to solve this equation for t. We can factor it to give:
t [31 – 4.9t) = 0
t = 31/4.9 = 6.4 s
The ball spends a total of 6.4 seconds in flight.
Problem #3
A ball is thrown directly downward with an initial speed of 8.00 m s from a height of 30.0 m. When does the ball strike the ground?
Answer:
We diagram the problem as in Figure. We have to choose a coordinate system, and here I will put the let the origin of the y axis be at the place where the ball starts its motion (at the top of the 30 m height). With this choice, the ball starts its motion at yi = 0 and strikes the ground when yf = −30 m.
We can now see that the problem is asking us: At what time does yf = −30.0 m? We have vi = −8.00 m/s (minus because the ball is thrown downward!) and the acceleration of the the ball is a = −g = −9.8 m/s2 , so at any time t the y coordinate is given by
yf = yi + vit - ½gt2
-30.0 m = 0 + (-8.00 m/s)t – ½ (9.8 m/s2)t2
4.9t2 + 8t – 30 = 0
which is just a quadratic equation in t. From our algebra courses we know how to solve this; the solutions are:
t = 1.78 s
Our answer is one of these . . . which one? Obviously the ball had to strike the ground at some positive value of t, so the answer is t = 1.78 s. The ball strikes the ground 1.78 s after being thrown.
Problem #4
A student throws a set of keys vertically upward to her sorority sister in a window 4.00 m above. The keys are caught 1.50 s later by the sister’s outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught?
We diagram the problem as in Figure. We have to choose a coordinate system, and here I will put the let the origin of the y axis be at the place where the ball starts its motion (at the top of the 30 m height). With this choice, the ball starts its motion at yi = 0 and strikes the ground when yf = −30 m.
We can now see that the problem is asking us: At what time does yf = −30.0 m? We have vi = −8.00 m/s (minus because the ball is thrown downward!) and the acceleration of the the ball is a = −g = −9.8 m/s2 , so at any time t the y coordinate is given by
yf = yi + vit - ½gt2
-30.0 m = 0 + (-8.00 m/s)t – ½ (9.8 m/s2)t2
4.9t2 + 8t – 30 = 0
which is just a quadratic equation in t. From our algebra courses we know how to solve this; the solutions are:
t = 1.78 s
Our answer is one of these . . . which one? Obviously the ball had to strike the ground at some positive value of t, so the answer is t = 1.78 s. The ball strikes the ground 1.78 s after being thrown.
Problem #4
A student throws a set of keys vertically upward to her sorority sister in a window 4.00 m above. The keys are caught 1.50 s later by the sister’s outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught?
Answer:
(a) We draw a simple picture of the problem; such a simple picture is given in Figure. Having a picture is important, but we should be careful not to put too much into the picture; the problem did not say that the keys were caught while they were going up or going down. For all we know at the moment, it could be either one!
We will put the origin of the y axis at the point where the keys were thrown. This simplifies things in that the initial y coordinate of the keys is yi = 0. Of course, since this is a problem about free–fall, we know the acceleration: a = −g = −9.80 m/s2 . What mathematical information does the problem give us? We are told that when t = 1.50 s, the y coordinate of the keys is yf = 4.00 m. Is this enough information to solve the problem? We write the equation for y(t):
yf = yi + vit - ½gt2
4.00 m = vi(1.50 s) – ½ (9.8 m/s2)(1.50 s)2
Now we can solve for vi. Rearrange this equation to get:
1.50vi = 4.00 + (4.9)(1.50)2 = 15.0
vi = 10.0 m/s
(b) We want to find the velocity of the keys at the time they were caught, that is, at t = 1.50 s. We know vi; the velocity of the keys at all times follows from Eq.
vf = vi – gt = 10.0 m/s – (9.8 m/s2)(1.50) = -4.68 m/s
So the velocity of the keys when they were caught was −4.68 m/s . Note that the keys had a negative velocity; this tells us that the keys were moving downward at the time they were caught!
Problem #5
A ball is thrown vertically upward from the ground with an initial speed of 15.0 m/s . (a) How long does it take the ball to reach its maximum altitude? (b) What is its maximum altitude? (c) Determine the velocity and acceleration of the ball at t = 2.00 s.
(a) We draw a simple picture of the problem; such a simple picture is given in Figure. Having a picture is important, but we should be careful not to put too much into the picture; the problem did not say that the keys were caught while they were going up or going down. For all we know at the moment, it could be either one!
We will put the origin of the y axis at the point where the keys were thrown. This simplifies things in that the initial y coordinate of the keys is yi = 0. Of course, since this is a problem about free–fall, we know the acceleration: a = −g = −9.80 m/s2 . What mathematical information does the problem give us? We are told that when t = 1.50 s, the y coordinate of the keys is yf = 4.00 m. Is this enough information to solve the problem? We write the equation for y(t):
yf = yi + vit - ½gt2
4.00 m = vi(1.50 s) – ½ (9.8 m/s2)(1.50 s)2
Now we can solve for vi. Rearrange this equation to get:
1.50vi = 4.00 + (4.9)(1.50)2 = 15.0
vi = 10.0 m/s
(b) We want to find the velocity of the keys at the time they were caught, that is, at t = 1.50 s. We know vi; the velocity of the keys at all times follows from Eq.
vf = vi – gt = 10.0 m/s – (9.8 m/s2)(1.50) = -4.68 m/s
So the velocity of the keys when they were caught was −4.68 m/s . Note that the keys had a negative velocity; this tells us that the keys were moving downward at the time they were caught!
Problem #5
A ball is thrown vertically upward from the ground with an initial speed of 15.0 m/s . (a) How long does it take the ball to reach its maximum altitude? (b) What is its maximum altitude? (c) Determine the velocity and acceleration of the ball at t = 2.00 s.
Answer:
(a) An illustration of the data given in this problem is given in Figure. We measure the coordinate y upward from the place where the ball is thrown so that yi = 0. The ball’s acceleration while in flight is a = −g = −9.80 m/s2 . We are given that vi = +15.0 m s . The ball is at maximum altitude when its (instantaneous) velocity vf is zero (it is neither going up nor going down) and we can use the expression for v to solve for t:
vf = vi – gt
Plug in the values for the top of the ball’s flight and get:
t = (0 – 15 m/s)/(-9.8 m/s2) = 1.53 s
The ball takes 1.53 s to reach maximum height.
(b) we use formulas to determine the magnitude of final position,
vf2 = vi2 - 2g(yf – yi)
02 = (15.0 m/s)2 - 2(9.8 m/s2)(yf – 0)
225 = 19.6yf
yf = 11.48 m
The ball reaches a maximum height of 11.48 m.
(c) At t = 2.00 s (that is, 2.0 seconds after the ball was thrown) we use
vf = vi – gt = 15 m/s – (9.9 m/s2)(2.00 s) = -4.60 m/s
so at t = 2.00 s the ball is on its way back down with a speed of 4.60 m/s . As for the next part, the acceleration of the ball is always equal to −9.80 m/s2 while it is in flight.
Problem #6
A baseball is hit such that it travels straight upward after being struck by the bat. A fan observes that it requires 3.00 s for the ball to reach its maximum height. Find (a) its initial velocity and (b) its maximum height. Ignore the effects of air resistance.
(a) An illustration of the data given in this problem is given in Figure. We measure the coordinate y upward from the place where the ball is thrown so that yi = 0. The ball’s acceleration while in flight is a = −g = −9.80 m/s2 . We are given that vi = +15.0 m s . The ball is at maximum altitude when its (instantaneous) velocity vf is zero (it is neither going up nor going down) and we can use the expression for v to solve for t:
vf = vi – gt
Plug in the values for the top of the ball’s flight and get:
t = (0 – 15 m/s)/(-9.8 m/s2) = 1.53 s
The ball takes 1.53 s to reach maximum height.
(b) we use formulas to determine the magnitude of final position,
vf2 = vi2 - 2g(yf – yi)
02 = (15.0 m/s)2 - 2(9.8 m/s2)(yf – 0)
225 = 19.6yf
yf = 11.48 m
The ball reaches a maximum height of 11.48 m.
(c) At t = 2.00 s (that is, 2.0 seconds after the ball was thrown) we use
vf = vi – gt = 15 m/s – (9.9 m/s2)(2.00 s) = -4.60 m/s
so at t = 2.00 s the ball is on its way back down with a speed of 4.60 m/s . As for the next part, the acceleration of the ball is always equal to −9.80 m/s2 while it is in flight.
Problem #6
A baseball is hit such that it travels straight upward after being struck by the bat. A fan observes that it requires 3.00 s for the ball to reach its maximum height. Find (a) its initial velocity and (b) its maximum height. Ignore the effects of air resistance.
Answer:
(a) An illustration of the data given in the problem is given in Figure. For the period from when the ball is hit to the time it reaches maximum height, we know the time interval, the acceleration (a = −g) and also the final velocity, since at maximum height the velocity of the ball is zero. Then Eq. vf = vi – gt gives us vi:
vf = vi – gt
0 = vi – (9.8 m/s2)(3.0 s)
vi = 29.4 m/s
The initial velocity of the ball was +29.4 m/s .
(b) To find the value of the maximum height, we need to find the value of the y coordinate at time t = 3.00 s. we use formulas to determine the magnitude of maximum height,
vf2 = vi2 - 2g(yf – yi)
02 = (29.4 m/s)2 – 2(9.8 m/s2)(yf – yi)
(yf – yi) = (29.4)2/19.6 = 44.1 m
The ball reached a maximum height of 44.1 m.
(a) An illustration of the data given in the problem is given in Figure. For the period from when the ball is hit to the time it reaches maximum height, we know the time interval, the acceleration (a = −g) and also the final velocity, since at maximum height the velocity of the ball is zero. Then Eq. vf = vi – gt gives us vi:
vf = vi – gt
0 = vi – (9.8 m/s2)(3.0 s)
vi = 29.4 m/s
The initial velocity of the ball was +29.4 m/s .
(b) To find the value of the maximum height, we need to find the value of the y coordinate at time t = 3.00 s. we use formulas to determine the magnitude of maximum height,
vf2 = vi2 - 2g(yf – yi)
02 = (29.4 m/s)2 – 2(9.8 m/s2)(yf – yi)
(yf – yi) = (29.4)2/19.6 = 44.1 m
The ball reached a maximum height of 44.1 m.
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