Question

Sam heaves a 16 lb shot straight upward, giving it a constant upward acceleration from rest of 30.0 m/s² for a height of 68.0 cm. He releases it at a height of 2.30 m above the ground. Ignore air resistance. What is the maximum height reached by the shot?

1) 0.68 m
2) 2.30 m
3) 2.98 m
4) 3.36 m

Answer 1

The maximum height reached by the shot is 2.30 m.

Step-by-step explanation:

To find the maximum height reached by the shot, we first need to find the time it takes for the shot to reach its maximum height. We can use the equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0 m/s, the acceleration is -30.0 m/s² (negative because it is directed opposite to the motion), and the final velocity at the maximum height is also 0 m/s. Solving for t, we get:

t = -vi / a

t = 0 / (-30.0)

t = 0 s

Since the time it takes for the shot to reach its maximum height is 0 seconds, this means that the maximum height is reached as soon as the shot is released. So, the maximum height is equal to the initial height, which is 2.30 m above the ground. Therefore, the correct answer is 2.30 m.