Question

n the figure, a stone is projected at a cliff of height h with an initial speed of 48.0 m/s directed at an angle ?₀ = 64.0° above the horizontal. The stone strikes at A, 5.26 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.

Answer 1

Data Given

Initial Velocity v₀ = 48 m/s , Angle = 64⁰, Time t = 48 m/s

Let us first calculate Horizontal and vertical components of the velocity

$v_{0,x}= v_{0}\cos \theta = (48 m/s )\cos 64^{0} = 21.04 m/s$

$v_{0,y}= v_{0}\sin \theta = (48 m/s )\sin 64^{0} = 43.14 m/s$

as well a_x = 0, and a_y = -9.8 m/s²

Part A) For the vertical motion

$h = v_{0,y}t+\frac{1}{2}a_{y}t^{2}$

$h =43.14 m/s\times 5.26 s-\frac{1}{2}\times 9.8m/s^{2}\times (5.26s)^{2} =91.34 m$

Part B) Speed of the stone just before the impact

$v_{x} = v_{0,x}+a_{x}t =21.04 m/s + 0\times 5.26 s = 21.04 m/s$

$v_{y} = v_{0,y}+a_{y}t =43.14 m/s -9.8 m/s^{2}\times 5.26 s = -8.41 m/s$

Now total speed will be

$v =\sqrt{ v_{x}^{2}+v_{y}^{2}} = \sqrt{ (21.04 m/s)^{2}+(-8.41m/s)^{2}} = 22.66 m/s$

Part C) The maximum height H reached above the ground will be

$H = \frac{v_{0,y}^{2}}{2g} = \frac{(43.14 m/s)^{2}}{2\times 9.8 m/s^{2}} = 94.95 m$