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A baseball player wants to hit a home run over the wall of a stadium. The player swings the baseball bat so that it hits the ball when it is at a height of 0.996 m above the ground. The ball flies off at an angle of 30° above the horizontal and at a speed of 36.2 m/s. What is the tallest wall that the player can clear (i.e., get the ball over) if the wall is 99.1 m away horizontally?

Answer :

A wall less than or equal to 14.7 m in height is the highest the player can clear.

How to calculate height?

Use the kinematic equations of motion to solve the problem. First, we can find the time it takes for the ball to travel 99.1 m horizontally:

d = vt

t = d / v

t = 99.1 m / 36.2 m/s

t ≈ 2.74 s

Now, use the vertical motion equation to find the maximum height the ball reaches:

y = yo + vot + (1/2)at²

where:

yo = 0.996 m (initial height)

vo = v sinθ = 36.2 m/s x sin(30°) ≈ 18.1 m/s (initial vertical velocity)

a = -9.81 m/s² (acceleration due to gravity, pointing downward)

t = 2.74 s (time of flight)

y = 0.996 m + 18.1 m/s x 2.74 s + (1/2) x (-9.81 m/s²) x (2.74 s)²

y ≈ 14.7 m

Therefore, the tallest wall the player can clear is a wall with a height less than or equal to 14.7 m.

Find out more on baseball here: https://brainly.com/question/29659060

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