Question

The roof of a two-story house makes an angle of 28° with the horizontal. A ball rolling down the roof rolls off the edge at a speed of 4.9 m/s. The distance to the ground from that point is 6.6 m. (a) How long is the ball in the air? s (b) How far from the base of the house does it land? m (c) What is its velocity just before landing? (Let upward be the positive y-direction.) x-component m/s y-component m/s

Answer 1

Calculate the vertical component of velocity of ball.

u_y = vsin28^o = -4.9sin28^o = -2.3 m/s

h = -6.6 m

So we can use h = u_yt + 1/2 gt²

-6.6 = -2.3t - (9.8/2)t²

=> 4.9t² + 2.3t - 6.6 = 0

so t = 0.949 s and t = -1.418 s

but we can discard the negative value.

therefore t = 0.949 s

b) The horizontal component of velocity is independent of the acceleration. Hence the x distance covered will be:

x = u_xt

=> x = 4.9cos28^o x 0.949 = 4.1m

c) The horizontal component of velocity remains unchanged, but the vertical component changes.

For t = 0.949 s, we can use v_y = u_y + gt

so v_y = -2.3 - 9.8(0.949) = -11.6 m/s

and v_x = u_x = 4.9cos28^o = 4.326 m/s

So the magnitude of velocity just before hitting the ground will be:

Please rate if it is correct and and help full