Answer:
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:
[tex]v_{fy} = v_{oy} -gt[/tex] Formula (1)
[tex]y= y_{o} +v_{oy} *t - \frac{1}{2} *g*t^{2}[/tex] Formula (2)
Where:
y : vertical position for any time t (m)
y₀ : Initial vertical position (m)
t : time in seconds (s)
[tex]v_{oy}[/tex] : Initial speed in y (m/s)
[tex]v_{fy}[/tex] : 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, [tex]v_{oy}[/tex] :
at maximum height : t= 3 s, [tex]v_{fy}[/tex] = 0
[tex]v_{fy} = v_{oy} -g*t[/tex]
[tex]0 = v_{oy} -9.8*3[/tex]
[tex]v_{oy} = 29.4 \frac{m}{s}[/tex]
We calculate the requested height using formula 2
t= 3s+2.5s = 5.5 s, [tex]v_{oy} = 29.4 \frac{m}{s}[/tex] , y₀ = 0
[tex]y= y_{o} +v_{oy} *t - \frac{1}{2} *g*t^{2}[/tex]
[tex]y= 0 +(29.4) *(5.5) - \frac{1}{2} *(9.8)*(5.5)^{2}[/tex]
y=13.475 m
