Question

A boy stands at the edge of a cliff and launches a rock upward at an angle of 45.0 degrees.The rock comes back down to the elevation where it was released 2.25 s later,then continues until it is seen to splash into the lake below 4.00 s after release.How far below the point of release is the lake surface? What horizontal distance from the point of release is the splash?

Answer 1

considering motion along vertical

time taken to reach max height, t = 2.25/2 = 1.125 s

at max height

vy = 0 using 1st equation of motion

vy = uy - gt

0 = uy - 9.8* 1.125

uy = 11.025 m/s

now, velocity with which the rock projected

uy = u sin 45

u = 11.025/ sin 45 = 15.59 m/s

again considering motion along vertical

h = h0 + uy*t - 0.5* gt^2

(h - ho) = 11.025* 4 - 0.5*9.8* 4^2

(h - ho) = - 34.3 m

(ho - h) = 34.3 m (Ans)

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motion along horizontal

R = u cos 45 * 4

R = 15.59* cos 45 *4

R = 44.1 m

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