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

Step-by-step 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=√(2gh)`

Net velocity at ground

`37=√((u\cos 40)^2+(v_y)^2)`

`37^2=(10\cos 40)^2+2* 9.8* h`

`h=(1310.317)/(2* 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=√(v_y'^2+2gh)`

`v_0=√((10\sin 40)+2* 9.8* 66.85)`

`v_0=36.76`

Final velocity at ground

`=√(v_0^2+u_x^2)`

`=√(36.76^2+(10\cos 40)^2)`

`=37.55 m/s`