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.

Question

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.

Answer 1

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)