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?

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:

 Formula (1)

 Formula (2)

Where:  

y :  vertical position for any time t  (m)  

y₀ : Initial vertical position   (m)

t :    time in seconds (s)

     : Initial speed in y (m/s)  

     : 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,  :

at maximum height : t= 3 s,    = 0

We calculate the requested height using formula 2

t= 3s+2.5s = 5.5 s, , y₀ = 0

y=13.475 m

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards