Question

Emily throws a soccer ball out of her dorm window to Allison, who is waiting below to catch it.

If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should Allison stand to catch the ball? Assume the vertical distance between where Emily releases the ball and Allison catches it is 8.0m.

Answer 1

Let us consider the downwards direction as positive.

Gravitational acceleration = g = 9.81 m/s²

Initial velocity of the ball = V = 14 m/s

Angle the ball is thrown at = = 30^o

Initial horizontal velocity of the ball = V_x = VCos = (14)Cos(30) = 12.12 m/s

Initial vertical velocity of the ball = V_y = VSin = (14)Sin(30) = 7 m/s

Vertical distance between where Emily throws the ball and Allison catches it = H = 8 m

Time period the ball is in air = T

H = V_yT + gT²/2

8 = (7)T + (9.81)T²/2

4.905T² + 7T - 8 = 0

T = 0.75 sec or -2.18 sec

Time cannot be negative.

T = 0.75 sec

Horizontal distance traveled by the ball = R

There is no horizontal force on the ball therefore the horizontal velocity of the ball remains constant.

R = V_xT

R = (12.12)(0.75)

R = 9.09 m

Distance of Allison from the base of the dorm to catch the ball = 9.09 m