Problem#1
An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15.0 m if her initial speed is 3.00 m/s. What is the free-fall acceleration on the planet?Answer:
Given:
maximum range, R = 15.0 m
initial speed, v = 3.00 m/s
and θmax = 45.00
then, the free-fall acceleration on the planet given by
R = vi2 sin 2θ/g
15.0 m = (3.00 m/s)2 sin (2 x 45.00)/g
g = 0.600 m/s2
Problem#2
A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection?
Answer:
the maximum height that the motion of the projectile is given by
h = vi2 sin2θi/2g
the maximum range of projectile is
R = vi2 sin 2θi/g
Because,
3h = R
3vi2 sin2θi/2g = vi2 sin 2θi/g
3 sin2θi/2 = (2 sin θi cos θi)
4/3 = tan θi
Thus, θi = tan-1(4/3) = 53.10
Problem#3
A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal range d. (a) At what angle ' is the rock thrown? (b) What If? Would your answer to part (a) be different on a different planet? (c) What is the range dmax the rock can attain if it is launched at the same speed but at the optimal angle for maximum range?
Answer:
(a) To identify the maximum height we let i be the launch point and f be the highest point:
vyf2 = vyi2 + 2ayΔy
0 = (vi cos θi)2 + 2(–g)(ymax – 0)
ymax = vi2 sin2θi/2g
To identify the range we let i be the launch and f be the impact point; where t is not zero:
yf = yi + vyit + ½ ayt2
0 = 0 + (vi sin θi)t + ½ (–g)t2
t = 2vi sin θi/g
and
xf = xi + vxit
d = 0 + (vi cos θi)(2vi sin θi/g)
For this rock, d = ymax
(vi cos θi)(2vi sin θi/g) = vi2 sin2θi/2g
sin θi/cos θi = 4
tan θi = 4
θi = tan-1(4) = 76.00
(b) Since g divides out, the answer is the same on every planet.
(c) The maximum range is attained for θi = 45° :
dmax/d = {(vi cos 450)(2vi sin 450)g}/{(gvi cos 760)(2vi sin 760)}
dmax/d = 2.125
so, dmax = 17d/8
Problem#4
A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0° below the horizontal. It strikes the ground 3.00 s later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the level of launching?
Answer:
(a) The horizontal distance traveled is
xf = xi + vxit = 0 + 8.00 m/s cos 20.00 x 3.00 s = 22.6 m
(b) the height from which the ball was thrown is
yf = yi + vyit + ½ ayt2
yf = yi + (–vi sin θi)t + ½ (–g)t2
0 = yi + (–8.00 m/s) sin 20.00 x 3.00 s + ½ (–9.80 m/s2)(3.00 s)2
yi = 52.3 m
(c) the length of time it takes for the ball to reach the point of 10.0 m below the launch rate is
yf = yi + vyit + ½ ayt2
yf – yi = vyit + ½ ayt2
– 10 m = (–8.00 m/s) sin 20.00 x t + ½ (–9.80 m/s2)(t)2
4.9t2 + 2.74t – 10 = 0
t = {–2.74 ± [(2.74)2 + 196]1/2}/9.80 = 1.18 s
Problem#5
The small archerfish (length 20 to 25 cm) lives in brackish waters of southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of 1.2 m to 1.5 m, and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target 2.00 m away, at an angle of 30.0o above the horizontal. With what velocity must the water stream be launched if it is not to drop more than 3.00 cm vertically on its path to the target?
Answer;
We interpret the problem to mean that the displacement from fish to bug is
2.00 m at 30°= 2.00 m cos30°i + 2.00 m sin30°j = 1.73 m i + 1.00 mj.
If the water should drop 0.03 m during its flight, then the fish must aim at a point 0.03 m above the
bug. The initial velocity of the water then is directed through the point with displacement
(1.73 m)i + (1.03 m)j = 2.015 m at 30.7°.
For the time of flight of a water drop we have
xf – xi = vxit
1.73 m – 0 = vi cos 30.70t
t = 1.73 m/vi cos 30.70
The vertical motion is described by
yf = yi + vyit + ½ ayt2
The “drop on its path” is
–3.00 cm = ½ (–9.80 m/s2)(1.73 m/vi cos 30.70)2
Thus,
vi = [1.73 m/cos 30.7][√[(9.80 m/s2)/(2 x 0.03 m)] = 25.8 m/s
xf = xi + vxit = 0 + 8.00 m/s cos 20.00 x 3.00 s = 22.6 m
(b) the height from which the ball was thrown is
yf = yi + vyit + ½ ayt2
yf = yi + (–vi sin θi)t + ½ (–g)t2
0 = yi + (–8.00 m/s) sin 20.00 x 3.00 s + ½ (–9.80 m/s2)(3.00 s)2
yi = 52.3 m
(c) the length of time it takes for the ball to reach the point of 10.0 m below the launch rate is
yf = yi + vyit + ½ ayt2
yf – yi = vyit + ½ ayt2
– 10 m = (–8.00 m/s) sin 20.00 x t + ½ (–9.80 m/s2)(t)2
4.9t2 + 2.74t – 10 = 0
t = {–2.74 ± [(2.74)2 + 196]1/2}/9.80 = 1.18 s
Problem#5
The small archerfish (length 20 to 25 cm) lives in brackish waters of southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of 1.2 m to 1.5 m, and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target 2.00 m away, at an angle of 30.0o above the horizontal. With what velocity must the water stream be launched if it is not to drop more than 3.00 cm vertically on its path to the target?
Answer;
We interpret the problem to mean that the displacement from fish to bug is
2.00 m at 30°= 2.00 m cos30°i + 2.00 m sin30°j = 1.73 m i + 1.00 mj.
If the water should drop 0.03 m during its flight, then the fish must aim at a point 0.03 m above the
bug. The initial velocity of the water then is directed through the point with displacement
(1.73 m)i + (1.03 m)j = 2.015 m at 30.7°.
For the time of flight of a water drop we have
xf – xi = vxit
1.73 m – 0 = vi cos 30.70t
t = 1.73 m/vi cos 30.70
The vertical motion is described by
yf = yi + vyit + ½ ayt2
The “drop on its path” is
–3.00 cm = ½ (–9.80 m/s2)(1.73 m/vi cos 30.70)2
Thus,
vi = [1.73 m/cos 30.7][√[(9.80 m/s2)/(2 x 0.03 m)] = 25.8 m/s
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