Question

A soccer player kicks the ball toward a goal that is 25.0 m in front of him. The ball leaves his foot at a speed of 17.0 m/s and an angle of 36.0 ° above the ground. Find the speed of the ball when the goalie catches it in front of the net.

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the ball = V₁ = 17 m/s

Angle the ball is kick at = = 36^o

Initial horizontal velocity of the ball = V_1x = V₁Cos = (17)Cos(36) = 13.75 m/s

Initial vertical velocity of the ball = V_1y = V₁Sin = (17)Sin(36) = 9.992 m/s

Distance of the ball from the goal = R = 25 m

Time taken by the ball to reach the goal = T

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

R = V_1xT

25 = (13.75)T

T = 1.818 sec

Speed of the ball when the ball reaches the goal = V₂

Horizontal velocity of the ball when it reaches the goal = V_2x

V_2x = V_1x

V_2x = 13.75 m/s

Vertical velocity of the ball when it reaches the goal = V_2y

V_2y = V_1y + gT

V_2y = 9.992 + (-9.81)(1.818)

V_2y = -7.843 m/s

V₂ = 15.8 m/s

Speed of the ball when the goalie catches it in front of the net = 15.8 m/s