Question

A soccer ball is kicked from the top of one building with a height of H₁ = 30.1 m to another building with a height of H₂ = 10.8 m. (It is not a very smart idea to play soccer on the roof of tall buildings.)

The ball is kicked with a speed of v₀ = 19.3 m/s at an angle of θ = 74.0° with respect to the horizontal. The mass of a size 5 soccer ball is m = 450 g. What is the speed of the soccer ball, when it lands on the roof of the second building? The soccer ball is kicked without a spin. Neglect air resistance.

Answer 1

Gravitational acceleration = g = -9.81 m/s²

Height of the first building = H₁ = 30.1 m

Height of the second building = H₂ = 10.8 m

Initial velocity of the ball = V₀ = 19.3 m/s

Angle of initial velocity with the horizontal = = 74^o

Initial horizontal velocity of the ball = V_x0

V_x0 = V₀Cos

V_x0 = (19.3)Cos(74)

V_x0 = 5.32 m/s

Initial vertical velocity of the ball = V_y0

V_y0 = V₀Sin

V_y0 = (19.3)Sin(74)

V_y0 = 18.552 m/s

Change in height of the ball = h

h = H₂ - H₁

h = 10.8 - 30.1

h = -19.3 m

Time of flight of the ball = T

h = V_y0T + gT²/2

-19.3 = (18.552)T + (-9.81)T²/2

4.905T² - 18.552T - 19.3 = 0

T = 4.632 or -0.849

Time cannot be negative.

T = 4.632 sec

Horizontal velocity of the ball when it lands on the second building = V_x1

There is no force acting on the ball in the horizontal direction hence the velocity of the ball in the horizontal direction remains constant.

V_x1 = V_x0

V_x1 = 5.32 m/s

Vertical velocity of the ball when it lands on the second building = V_y1

V_y1 = V_y0 + gT

V_y1 = 18.552 + (-9.81)(4.632)

V_y1 = -26.888 m/s

Speed of the ball when it lands on the second building = V₁

V₁ = 27.41 m/s

Speed of the soccer ball when it lands on the roof of the second building = 27.41 m/s