Asked by j
If he leaves the ramp with a speed of 36.5 m/s and has a speed of 35.0 m/s at the top of his trajectory, determine his maximum height (h) (in m) above the end of the ramp. Ignore friction and air resistance.
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Answered by
GPT-5 mini
AI
Use energy (or kinematics). The loss of kinetic energy from launch to the top goes into gravitational potential:
mgh = 1/2 m (v0^2 − v_top^2) => h = (v0^2 − v_top^2)/(2g).
Plugging in v0 = 36.5 m/s, v_top = 35.0 m/s, g = 9.80 m/s^2:
h = (36.5^2 − 35.0^2)/(2·9.80) = (1332.25 − 1225)/19.6 = 107.25/19.6 ≈ 5.47 m.
So the maximum height above the ramp is about 5.47 m.
mgh = 1/2 m (v0^2 − v_top^2) => h = (v0^2 − v_top^2)/(2g).
Plugging in v0 = 36.5 m/s, v_top = 35.0 m/s, g = 9.80 m/s^2:
h = (36.5^2 − 35.0^2)/(2·9.80) = (1332.25 − 1225)/19.6 = 107.25/19.6 ≈ 5.47 m.
So the maximum height above the ramp is about 5.47 m.
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