Look out below! Physics problem?

Question

Sam heaves a shot with weight 16.0 lbs straight upward, giving it a constant upward acceleration from rest of 45.2 m/s^2 for a height 65.0 m . He releases it at height 2.14 m above the ground. You may ignore air resistance.

Part A: What is the speed of the shot when he releases it?

Part B: How high above the ground does it go?

Part C: How much time does he have to get out of its way before it returns to the height of the top of his head, a distance 1.84 m above the ground?

Take the free fall acceleration to be g = 9.80 m/s^2 for all parts of this problem.

Thanks guys, I really appreciate the help :D

Answer 1

First of all.... 16lbs/2.2kg/lb = 7.27 kg.

A)
Since it travels a total distance of 65-2.14=62.86 m, remember that v²=vi²-2gy and, at the top, v=0 so
vi=√(2*9.8*62.86)=35.1 m/s

B)
Duh..... You -told- us it went to a height of 65 m. Did you lie about it? ☺

C)
From 65 m to 1.84 m is 63.16 m. since s=at²/2 then
t=√(2s/a)=√(2*63.16/9.8)=3.56 s

Hope that helps ☺


Doug