Question

In the figure, a stone is projected at a cliff of height h with an initial speed of 47.0 m/s directed at an angle θ₀ = 54.0° above the horizontal. The stone strikes at A, 5.48 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c)the maximum height H reached above the ground. Use g=9.80 m/s².

Chapter 04, Problem 028 In the figure, a stone is projected at a cliff of height h with an initial speed of 47.0 m/s directed at an angle θ。= 54.0° above the horizontal. The stone strikes at A, 5.48 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground. Use g- 9.80 m/s 2

Answer 1

a] Using second equation of motion in vertical,

H = u sin theta t - 0.5 gt^2

= 47*sin 54 degree *5.48 - 4.9*5.48^2

= 61.22 m answer

b] By energy conservation,

Final KE - Initial KE = -mgh

0.5mv^2 - 0.5mu^2 = -mgh

v^2 -47^2 = 2*-9.8*61.22

v = sqrt(47^2 + 2*-9.8*61.22)

= 31.77 m/s

c] Maximum height H = [u sin theta]^2/2g = [47* sin 54 degree]^2/19.6

= 73.77 m