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Lelu [443]
3 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]3 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]3 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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a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

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a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

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Here

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                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

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  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

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P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

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