Physics: Abaseball 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?

908 KPaExplanation:See it in the pictures....

Abaseball is hit at ground level. the ball is observed

loxtin495

06.01.2022

y= 13.475 m

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)

$y= y_{o} +v_{oy} *t - \frac{1}{2} *g*t^{2}$ 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

$v_{fy} = v_{oy} -g*t$

$0 = v_{oy} -9.8*3$

$v_{oy} = 29.4 \frac{m}{s}$

We calculate the requested height using formula 2

t= 3s+2.5s = 5.5 s, $v_{oy} = 29.4 \frac{m}{s}$ , y₀ = 0

$y= y_{o} +v_{oy} *t - \frac{1}{2} *g*t^{2}$

$y= 0 +(29.4) *(5.5) - \frac{1}{2} *(9.8)*(5.5)^{2}$

y=13.475 m

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