I can\'t figure out how to do this problem since the angle is 30 and not 45...th

Question

I can't figure out how to do this problem since the angle is 30 and not 45...the components of velocity would not be equal and I'm lost A ball of mass m is dropped at a height of 2h above a frictionless plane inclined at an angle of 30 to the horizontal as shown. At point A, the ball bounces elastically off the plane (horizontally with no loss of energy or velocity) and then strikes the plane again at point B. In terms of g and h, determine each of the following: a) The time it takes the mass to fall 2h b) The velocity of the mass as it bounces off the plane at point A c) The time it takes to travel between points A and B d) The distance 4Lalong the plane between points A and B e) The velocity of the mass as it strikes the plane at point B las 242h. Q m 2h Y 30

Explanation / Answer

(A) displacement as it falls by 2h distance,

y = v0t + a t^2 /2

- 2 h = 0 - g t^2 /2

t = sqrt(4 h / g) = 2 sqrt(h/g)

(B) v = a t = g 2 sqrt(h/g) = 2 sqrt(hg)

(c) perpendicular to the plane,

v0 = v sin30 = sqrt(hg)

a = - g cos30

displacement from A to B along perpendicular direction of incline is zero.

y = v0t + a t^2 /2

0 = sqrt(hg) t - g cos30 t^2 / 2

t = 2 sqrt(h g) / g cos30

t = 2.31 sqrt(h/g)

(d) along the incline,

v0 = v cos30

a = g sin30

t = 2.31 sqrt(h/g)

4L = (2 sqrt(hg) cos30 2.31 sqrt(h/g)) + (g sin30) (2.31 sqrt(h/g))^2 /2

4L = 4 h + 13.07 h

4L = 17.07 h ..........Ans

(e) m g (2h + 4 L sin30) = m v^2 /2

2g (2h + 8.54h) = v^2

v = sqrt(21.07 g h)

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