Question

A baseball is hit at ground level. The ball is observed to reach its maximum height above ground level 3.0 s after being hit. And 2.5 s after reaching this maximum height, the ball is observed to barely clear a fence that is 97.5 m from where it was hit. How high is the fence?

Answer 1

y= 13.475 m

Step-by-step explanation:

Kinematics of the ball : The ball describes a parabolic trajectory.

Because ball move with uniformly accelerated movement in y we apply the following formulas:

`v_(fy) = v_(oy) -gt` Formula (1)

`y= y_(o) +v_(oy) *t - (1)/(2) *g*t^(2)` Formula (2)

Where:

y : vertical position for any time t (m)

y₀ : Initial vertical position (m)

t : time in seconds (s)

`v_(oy)` : Initial speed in y (m/s)

`v_(fy)` : Final speed in y (m/s)

g: acceleration due to gravity (m/s)²

Problem development

We apply formula 1 when the ball reaches its maximum height to obtain the initial velocity at y,
`v_(oy)` :

at maximum height : t= 3 s,
`v_(fy)` = 0

`v_(fy) = v_(oy) -g*t`

`0 = v_(oy) -9.8*3`

`v_(oy) = 29.4 (m)/(s)`

We calculate the requested height using formula 2

t= 3s+2.5s = 5.5 s,
`v_(oy) = 29.4 (m)/(s)` , y₀ = 0

`y= y_(o) +v_(oy) *t - (1)/(2) *g*t^(2)`

`y= 0 +(29.4) *(5.5) - (1)/(2) *(9.8)*(5.5)^(2)`

y=13.475 m