Answer:
a)Vg=13.42m/s :Speed with which the ball hits the ground
b) t₁= 2.74s : Time the ball remains in the air
c)h=9.19m: Maximum height reached by the ball
Explanation:
We apply the kinematic equations of parabolic motion:
a) Vg= Vo
Vg:speed with which the ball hits the ground
Vo: initial speed
Initial Speed Calculation
[tex]v_{o} =\sqrt{v_{ox}^{2} +v_{oy} ^{2} }[/tex]
[tex]v_{o} =\sqrt{16^{2} +12^{2} }[/tex]
Vo=13.42m/s
Vg=13.42m/s
b)Calculation of the time the ball remains in the air
t₁=2*t₂
t₁;time the ball remains in the air
t₂ time when the ball reaches the maximum height
Vf=Vo-g*t₂ : When the ball reaches the maximum height Vf = 0
0=13.42-9.8*t₂
9.8*t₂=13.42
t₂=13.42 ÷9.8
t₂=1.37s
t₁=2*1.37s
t₁= 2.74s
c)Calculation of the maximum height reached by the ball
When the ball reaches the maximum height Vf = 0
Vf²=V₀²-2*g*h
0= V₀²-2*g*h
2*g*h= V₀²
h= V₀² ÷ 2*g
h= 13.42² ÷2*9.8
h=9.19m
