A hard steel ball is dropped from rest at postion y = 2h and bounces from a hard

Question

A hard steel ball is dropped from rest at             postion y = 2h and bounces from a hard surface tilted at an angle             phi from horizontal. On impact, the ball's component of velocity             perpendicular to the surface is reversed (equal magntude oppostite             direction), while the velocity component parallel to the surface is             unchanged. (This is called a perfectly elastic collison.) Find the             horizontal displacment of the ball when it hits the ground             (at y =                                                                                                                                                                                                 0) from the point of impact with the surface.

Explanation / Answer

let t is the time taken to fall

y = 0.5*g*t^2

2*h = 0.5*9.8*t^2

==> t = sqrt(4*h/9.8) = 0.64*h s

x = 2*t*vox

here vox is the initial velocity in x direction

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