Problem#1
A soccer player kicks a rock horizontally off a 40.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.Answer:
The horizontal kick gives zero vertical velocity to the rock. Then its time of flight follows from
yf = yi + vyit + ½ ayt2
0 = 40.0 m + 0 + ½ (–9.80 m/s2)t2
t = 2.86 s
The extra time 3.00 s – 2.86 s = 0.143 s is the time required for the sound she hears to travel straight back to the player. It covers distance
(343 m/s)(0.143 s) = √[x2 + (40.0 m)2]
49.0 m = √[x2 + (40.0 m)2]
x = 28.3 m
where x represents the horizontal distance the rock travels.
xf = xi + vxit
28.3 m = 0 + vxi(2.86 s)
Then
vxi = 9.91 m/s
Problem#2
A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. 1). His motion through space can be modeled precisely as that of a particle at his center of mass, which we will define in Chapter 9. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.85 m above the floor, and is at elevation 0.900 m when he touches down again. Determine (a) his time of flight (his “hang time”), (b) his horizontal and (c) vertical velocity components at the instant of takeoff, and (d) his takeoff angle. (e) For comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations yi = 1.20 m, ymax = 2.50 m, yf = 0.700 m.
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| Fig.1 |
Answer:
From the instant he leaves the floor until just before he lands, the basketball star is a projectile. His vertical velocity and vertical displacement are related by the equation
vyf2 = vyi2 + 2ay(yf - yi)
02 = vyi2 + 2(–9.80 m/s2)(1.85 m – 1.02 m)
From this, vyi = 4.03 m/s
For the downward part of the flight, the equation gives
vyf2 = 0 + 2(–9.80 m/s2)(0.900 m – 1.85 m)
Thus the vertical velocity just before he lands is
vyf = –4.32 m/s
(a) His hang time may then be found from
vyf = vyi + (–g)t
–4.32 m/s = 4.03 m/s + (–9.80 m/s2)t
t = 0.852 s
(b) Looking at the total horizontal displacement during the leap,
xf = xi + vxt, becomes
2.80 m = vxi(0.852 s)
vxi = 3.29 m/s
(c) vertical velocity components at the instant of takeoff, vyi = 4.03 m/s. See above for proof.
(d) The takeoff angle is:
tan θ = (vyi/vxi)
tan θ = (4.03/3.29)
θ = tan-1(4.03/3.29) = 50.80
(e) Similarly for the deer, the upward part of the flight gives
vyf2 = vyi2 + 2ay (yf - yi)
0 = vyi2 + 2(–9.80 m/s2)(2.50 m – 1.20 m)
vyi = 5.04 m/s
For the downward part
vyf2 = vyi2 + 2ay (yf - yi)
vyf2 = 0 + 2(–9.80 m/s2)(0.700 m – 2.50 m)
vyf = –5.94 m/s
The hang time is then found as
vyf = vyi + ayt
–5.94 m/s = 5.04 m/s + (–9.80 m/s2)t
t = 1.12 s
Problem#3
An archer shoots an arrow with a velocity of 45.0 m/s at an angle of 50.0° with the horizontal. An assistant standing on the level ground 150 m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow. (a) What is the initial speed of the apple? (b) At what time after the arrow launch should the apple be thrown so that the arrow hits the apple?
Answer:
The arrow’s flight time to the collision point is
xf = xi + vxt
150 m = 0 + (45 m/s)t
t = 5.19 s
The arrow’s altitude at the collision is
yf = yi + vyit + ½ ayt2
yf = 0 + (45 m/s)(sin 500)(5.19 s) + ½ (–9.80 m/s2)(5.19 s)
yf = 47.0 m
(a) The required launch speed for the apple is given by
vyf2 = vyi2 + 2ay (yf - yi)
0 = vyi2 + 2(–9.80 m/s2)(47.0 m – 0)
vyi = 30.3 m/s
(b) The time of flight of the apple is given by
vyf = vyi + ayt
0 = 30.3 m/s + (–9.80 m/s2)t
t = 3.10 s
So the apple should be launched after the arrow by
5.19 s – 3.10 s = 2.09 s
Problem#4
A fireworks rocket explodes at height h, the peak of its vertical trajectory. It throws out burning fragments in all directions, but all at the same speed v. Pellets of solidified metal fall to the ground without air resistance. Find the smallest angle that the final velocity of an impacting fragment makes with the horizontal.
Answer:
For the smallest impact angle
tan θi = (vyf/vyi)
we want to minimize vyf and maximize vxf = vxi.
The final y-component of velocity is related to vyi by
vyf2 = vyi2 + 2ay (yf - yi) = vyi2 + 2gh
so we want to minimize vyi and maximize vxi . Both are accomplished by making the initial velocity
horizontal. Then vxi = v, vyi = 0, and
vyf = √(2gh)
At last, the impact angle is
θi = tan-1 [√(2gh)/v]
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