Question

A player kicks a ball with an initial vertical velocity of 12 m/s and horizontal velocity of 16 m/s. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?

Answer 1

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

Step-by-step 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

`v_(o) =\sqrt{v_(ox)^(2) +v_(oy) ^(2)  }`

`v_(o) =\sqrt{16^(2) +12^(2)  }`

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