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A ball player catches a ball 3.37s after throwing it vertically upward. With what speed did he throw it? What height did it reach?

User Marleen
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Answer: The initial velocity with which the ball was thrown is approximately 16.5 m/s. and the ball reached a height of approximately 13.9 meters.

Explanation: To find the initial velocity with which the ball was thrown and the height it reached, we can use the kinematic equations of motion. Let's assume the upward direction as positive.

Finding the initial velocity (u):

The time taken to reach the maximum height is equal to the time taken to return to the initial position. In this case, the ball player catches the ball after 3.37 seconds. Therefore, the time taken to reach the maximum height is half of that time:

t_up = 3.37s / 2 = 1.685s

Using the kinematic equation:

v = u + at

Since the ball reaches the maximum height, the final velocity (v) at that point is zero (as the ball momentarily stops). The acceleration (a) due to gravity is approximately -9.8 m/s² (taking downward as negative).

0 = u + (-9.8 m/s²) * (1.685s)

u = 9.8 m/s * 1.685s

u ≈ 16.5 m/s

Therefore, the initial velocity with which the ball was thrown is approximately 16.5 m/s.

Finding the height reached:

We can use the following kinematic equation to find the height (h):

v^2 = u^2 + 2ah

Since the final velocity (v) is zero at the maximum height, the equation becomes:

0 = (16.5 m/s)^2 + 2 * (-9.8 m/s²) * h

Simplifying the equation:

0 = 272.25 m²/s² - 19.6 m/s² * h

19.6 m/s² * h = 272.25 m²/s²

h = 272.25 m²/s² / 19.6 m/s²

h ≈ 13.9 meters

Therefore, the ball reached a height of approximately 13.9 meters.

User Vadim Kiselev
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