Physics
31.10.2021 09:27
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2. a slingshot is used to launch a stone horizontally from the top of a 200 meter

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?
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connormaxie
connormaxie
4,4(10 marks)

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+\frac{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{\frac{2s}{g}}=\sqrt{\frac{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 = \frac{d}{t}=\frac{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=\sqrt{v_x^2+v_y^2}=\sqrt{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}(\frac{v_y}{v_x})=tan^{-1}(\frac{62.6}{5.6})=84.9^{\circ}

Tyrant4life
Tyrant4life
4,9(82 marks)
When splitting an atom, a neutron, traveling at the right speed, is shot at the nucleus. usually the nucleus will split into two pieces and energy is released. also known as nuclear fission.

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