Chapter 5, Problem 68 Flying Circus of Physics A shot putter launches a 7.25 kg

Question

Chapter 5, Problem 68


Flying Circus of Physics

A shot putter launches a 7.25 kg shot by pushing it along a straight line of length 1.45 m and at an angle of 34.4° from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.3 m/s (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of 2.16 m and at an angle of 34.4°, and it lands at a horizontal distance of 16.0 m.What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)

Explanation / Answer

h = vy t - 1/2 g t^2 distance shot falls
t = s / vx time shot is in air
h = s vy / vx - 1/2 g (s / vx)^2 substituting for t
vy / vx = tan theta
vx^2 = 1/2 * g s^2 / (s tan theta - h)
vx^2 = 1/2 * 9.8 * 16^2 / (16 * tan 34.4 + 2.16) = 95.64
vx = 9.78 m/s when shot leaves ramp
t = 16 / 9.78 = 1.64 sec
vy = vx * tan 34.4 = 6.70 m/s
v =( 9.78^2 + 6.7^2)^1/2 = 11.8 m/s
V(avg) = 11.8 / 2 = 5.93 m/s average speed of shot on ramp
T = 1.45 / 5.93 = .245 sec time shot is on ramp
a = v / t = 11.8 / .245 = 48.2 m/s^2
F = m a + m g sin 34.4 = m (a + g sin 34.4) force along ramp
Substitute and solve

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