Question

Irina standing on the edge of a cliff throws a stone at 40 degree angle above the horizontal with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 37 m/s. If Irina were to throw the other rock at 40 degree angle below the horizontal from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

Answer 1

Answer:

Explanation:

Given

First stone is thrown at an angle of $\theta =40^{\circ}$ Above horizontal

second stone is thrown at angle of $40^{\circ}$ below horizontal

let h be the height of building

Initial velocity is same in both case i.e. u=10 m/s

Horizontal velocity will remain same as there is no acceleration and there will only be vertical velocity change

for first case

$v_y^2-v^2=2gh$

$v_y=\sqrt{2gh}$

Net velocity at ground

$37=\sqrt{(u\cos 40)^2+(v_y)^2}$

$37^2=(10\cos 40)^2+2\times 9.8\times h$

$h=\frac{1310.317}{2\times 9.8}=66.85 m$

For second case

$u_x=u\cos 40$

$v_y'=u\sin 40$

Let $v_0$ be the final vertical velocity

$v_0^2-v_y'^2=2gh$

$v_0=\sqrt{v_y'^2+2gh}$

$v_0=\sqrt{(10\sin 40)+2\times 9.8\times 66.85}$

$v_0=36.76$

Final velocity at ground

$=\sqrt{v_0^2+u_x^2}$

$=\sqrt{36.76^2+(10\cos 40)^2}$

$=37.55 m/s$