Question

2. A slingshot is used to launch a stone horizontally from the top of a 200 meter litt. The stone

lands 36.0 meters away.
a) At what speed was the stone launched?
b) What is the speed and angle of impact?

Answer 1

a) 5.6 m/s

First of all, we consider the vertical motion of the stone, which is a free fall motion (uniform accelerated motion). We can use the suvat equation :

`s=ut+(1)/(2)at^2`

where

s = 200 m is the vertical displacement (here we chose downward as positive direction)

u = 0 is the initial vertical velocity of the stone (since it is launched horizontally)

t is the time

`a=g=9.8 m/s^2` is the acceleration of gravity

Solving for t, we find the time the stone takes to reach the ground:

`t=\sqrt{(2s)/(g)}=\sqrt{(2(200))/(9.8)}=6.39 s`

Now we can consider the horizontal motion: this is a uniform motion with constant speed. The horizontal distance travelled is given by

`d=v_x t`

where

d = 36.0 m is the horizontal distance travelled by the stone

t = 6.39 s is the time of flight

If we solve for
`v_x`, we find the speed at which the stone was launched (which remains constant during the whole motion):

`v_x = (d)/(t)=(36)/(6.39)=5.6 m/s`

(b) 62.8 m/s at
`84.9^(\circ)` below the horizontal

- The motion along the horizontal direction is a uniform motion, since there are no forces acting on the stone in this direction - so the horizontal velocity remains constant:

`v_x = 5.6 m/s`

- The vertical velocity instead changes due to the effect of the acceleration of gravity. We can calculate the vertical velocity at the time of impact by using the equation

`v_y = u_y + gt`

where

`u_y = 0` is the initial vertical velocity (zero because the stone is launched horizontally)

Solving for t = 6.39 s, we find:

`v_y = 0+(9.8)(6.39)=62.6 m/s`

Keep in mind that the direction of this velocity is downward.

So now we can find the speed of the stone at the moment of impact:

`v=√(v_x^2+v_y^2)=√(5.6^2+62.6^2)=62.8 m/s`

And the angle of impact, measured as below the horizontal (since the vertical velocity is downward) is

`\theta=tan^(-1)((v_y)/(v_x))=tan^(-1)((62.6)/(5.6))=84.9^(\circ)`