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Lelu [443]
2 years ago
5

A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.

Physics
2 answers:
alisha [4.7K]2 years ago
7 0
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.

9966 [12]2 years ago
3 0

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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1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

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x=\frac{F}{k}=\frac{33 N}{200 N/m}=0.165 m

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