Physics, 02.10.2019 15:30 madison1284

Aslingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. 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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A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s$

Through Definition of Velocity, comes:

$\Delta v= \frac{\Delta S}{\Delta t} \\ v_x= \frac{36}{2} \\ \boxed {v_{x}=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}$

The angle of impact is given by:

$cos(\theta) =\frac{v_{x}}{v_{y}} \\ cos(\theta) = \frac{18}{20} \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.

Aslingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. the stone land

Aslingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. the stone land

Answer from: maddie02294

As per the question a stone is projected horizontally from the top of a 20 meter cliff.

Hence the height [h] of the cliff = 20 m.

The stone lands 36.0 m away.

Hence horizontal distance R] = 36 m.

First we are asked to calculate the launching speed of the stone.

Let it be denoted as u.

The motion of particle is the force of gravity only.Hence it is a projectile motion.

The total time taken by a projectile when it is fired horizontally at a height h from the ground is -

$T=\sqrt{\frac{2h}{g} }$ [ where g is the acceleration due to gravity = 9.8 m/s^2 or 10 m/s^2]

$=\sqrt{\frac{2*20}{10} } s$

= 2 s

The range of the projectile will be-

$R= \ horizontal\ speed*\ total\ time\ of\ flight$

The horizontal speed for a projectile is uniform through out the motion of the projectile as long as gravity is constant.

In this situation the horizontal velocity is equal to the initial projected speed i.e u.

Hence horizontal distance R = u ×T

Putting the value or R and T in the above formula we get-

$R=u*\sqrt{\frac{2h}{g} }$

$36 = u*\sqrt{\frac{2*20}{10} }$

36 = u*2

$u =\frac{36}{2}$

u =18 m/s

Now we are asked to calculate the angle of impact.

Initially the vertical velocity of the particle is 0. The path of the trajectory of the particle will be parabolic due to the force of gravity.At any instant of time the horizontal component of the instantaneous velocity of the of the particle will be constant .

Let the horizontal component is denoted as $V_{x} \ where \ V_{x} = u$

The vertical component is calculated as follows-

We know that v = u +at

Here v is the final velocity, u is the initial speed, a is the acceleration and t is the instantaneous time.

For vertically downward motion under gravity u = 0 and a = -g

Let the vertical velocity at any instant is denoted as $V_{y}$

Hence

$V_{y} = 0 -gt$

$V_{y} = gt$ [Here we are taking only the magnitude]

Let the resultant velocity makes an angle α with the horizontal when it strikes the ground.

The horizontal and vertical component is denoted in the diagram below-

From the figure we see that -

$tan\alpha =\frac{V_{y} }{V_{x} }$

$tan\alpha =\frac{gt}{u}$

$tan\alpha =\frac{10*2}{18}$

$tan\alpha =1.11111$

$\alpha =48.01^0$ [ans]

Hence the angle of impact is close to 48 .1 degree .

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