Using impact and conservation of linear moment, coefficient of restitution and p

Question

Using impact and conservation of linear moment, coefficient of restitution and projectile motion equations when necessary, Solve the dynamics question.

A girl throws a ball at an inclined wall from a height of 3 ft, haitting the wall at A with a horizontal velocity v0 of 25 ft/s. The coefficient of restitution between the ball and the wall is 0.9 and there is no friction. Determine the distance d from the foot of the wall to point B where the ball will hit the ground after bouncing off the wall

Explanation / Answer

incoming velocity = 25 (i) + 0 (j)

velocity in direction perpendicular to the wall is : 25*cos(30) which is equal to 21.7

applying coefficient of restitution: we have velocity after collision in the direction perpendicular to the incline is : 21.7*0.9 = 19.5ft/s

now lets consider the point of collision as the origin; and we have a problem in projectile motion where the ball is thrown at an angle of 30 degrees from the horizontal from a height of 3 feet and we have to find the final position on the ground. so , the velocity is : -19.5*cos30 (i) + 19.5*sin30 (j) which is equal to : -16.9 (i) + 9.75 (j)

applying laws of motion equations in the vertical direction: -3 = 9.75t + 0.5*(-32)*t*t

solving this quadritc equation we get value of t as : 0.834 and -0.225; since we cannot have a negative value, we have time of flight as 0.834 seconds. so horizontal distance covered is 0.834*16.9 = 14.1 feet

but we have to subtract the horizontal distance of the origin from the base of the wall which is : 3/tan(60) which is 1.732

so distance between base of wall and the place where the ball drops on the ground is : 14.1-1.732 = 12.37 feet

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