a foul ball is hit straight up into the air with a speed of 32m/s and air resist

Question

a foul ball is hit straight up into the air with a speed of 32m/s and air resistance is negligible Part A Calculate the time required for the ball to rise to its maximum height. Part B Calculate the maximum height reached by the ball above the point where it hit the bat. Part C Determine the times at which the ball passes a point 28m above the point where it was hit by the bat a foul ball is hit straight up into the air with a speed of 32m/s and air resistance is negligible Part A Calculate the time required for the ball to rise to its maximum height. Part B Calculate the maximum height reached by the ball above the point where it hit the bat. Part C Determine the times at which the ball passes a point 28m above the point where it was hit by the bat a foul ball is hit straight up into the air with a speed of 32m/s and air resistance is negligible Part A Calculate the time required for the ball to rise to its maximum height. Part B Calculate the maximum height reached by the ball above the point where it hit the bat. Part C Determine the times at which the ball passes a point 28m above the point where it was hit by the bat

Explanation / Answer

H = Vit - 1/2 gt^2

where
Vi = the initial velocity
g= gravity
t = time
H = vertical distance or height

H = 32(t) - (1/2)(9.8)(t^2)

height is at maximum slope of the trajectory is 0. that is at the point when the ball is between rising and falling

The slope is the first derivative of the formula

Slope = dH = H' = 32 - 2(1/2)(9.8)t

Remember slope is 0 at maximum ht.
Therefore dH or H' = 0
plug it into the slope equation and solve for time (t)
0 = 32 - 9.8t
t = 32/9.8 sec = 3.26 s (Answer Part A)

Plug it in to the H equation (the first equation)
H = 32(3.26) - (1/2)(9.8)(3.26)^2

H = 52.24 m high (Answer Part B)

For part C, put H = 28m in above equation and solve for t.

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