Help with a physics question! When Babe Ruth hit a homer over the 6.0 m -high ri

Question

Help with a physics question!

When Babe Ruth hit a homer over the 6.0 m -high right-field fence 90 m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0 m above the ground and its path initially made a 45 angle with the ground.

I got 29.71 m/s the first time but the program didn0t accept it and responded with: Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures.

Explanation / Answer

Let the speed of ball be v.
Initial Velocity in x direction = v cos(45)
Initial Velocity in y direction = v sin(45)

Distance need to cover in horizontal direction = 90m
Distance need to cover in Vertical direction = 6m - 1m = 5m

Acceleration in horizontal direction = 0
Acceleration in Vertical direction = 9.8 m/s^2 downwards

Now time taken to cover horizontal distance t = 90/ v cos(45)

This is the same time needed to cover vertical distance 5m =

s = u*t - 0.5 * g * t^2
5 = v sin(45) * 90/ v cos(45) - 4.9 * (90/ v cos(45))^2
5 = 90 - 4.9 * (90/ v cos(45))^2
(90/ v cos(45)) = sqrt(85/4.9)
v cos(45) = 90/4.165
v = 90 / (4.165 * cos(45))
v = 30.55 m/s

Speed of ball when it left the bat v = 30.55 m/s

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