At the moment when a shot putter releases a 5.00 kg shot, the shot is 2.00 m above the ground and travelling at 15.0 m/s. It reaches a maximum height of 8.0 m above the ground and then falls to the ground. Assume air resistance is negligible.
a) What was the velocity of the shot at its maximum height?
I get this and can you check my answer of 10.36 m/s.
b) What was the velocity of the shot just before it strikes the ground?
I don't get how to do this.
c) As the shot comes to rest on the ground it creates a divet of 3.0 cm. What is the retarding force of the ground?
I also don't get this.
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1 answer

goes up from 2 to 8 or 6 meters
do vertical problem
v = Vi  g t
0 = Vi  9.8 t at the top (8meters)
so
t at top = Vi/9.8
h = 2 + Vi t  4.9 t^2
at top
8 = 2 + Vi t (1/2)(9.8) t^2
6 = Vi t (1/2)(9.8) t^2
6 = Vi^2/9.8  (1/2) Vi^2/9.8
12 (9.8) = Vi^2
Vi = 10.84 m/s
now do the combined vertical and horizontal
initial
sin A = 10.84/15
so A = angle up from horizontal = 46.3 degrees
horizontal speed = u = 15 cos 46.3 = 10.37 m/s forever (agree with you)
b) v = Vi  9.8 t
h = 0 at ground
0 = 2 + 10.84 t  4.9 t^2
4.9 t^2  10.84 t  2 = 0
t = [ 10.84 +/ sqrt (118+39.2) ]/9.8
t = 2.39 seconds to ground
so
v = 10.84  9.8(2.39) =  12.5 m/s
u is still 10.37
so speed = sqrt(12.5^2+10.^2)
tan angle below hor = 12.5/10.4
c) just vertical problem. It does not say how long the hole is
Force * time = change of momentum
F (distance/average speed) = 5(12.5)
average speed = 12.5/2
F (.03/6.25) = 5(12.5) 👍
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answered by Damon
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