Question

In: Physics

A soccer player kicks the ball toward a goal that is 25.0 m in front of...

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.

Expert Solution

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

genius_generous answered 1 year ago

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