Please answer the following questions: A rock is thrown horizontally with a spee

Question

Please answer the following questions:

A rock is thrown horizontally with a speed of 18 m/s from a vertical cliff of height 31 m. How long does it take to reach the horizontal ground below? How far will it land from the base of the cliff? What is the velocity (magnitude and direction counterclockwise from the +x-axis, which is the initial horizontal direction in which the rock was thrown) of the rock just before it hits the ground? A soccer ball is kicked with an initial speed of 28 m/s at an angle of 23 degree with respect to the horizontal, Find the maximum height reached by the ball, Find the speed of the ball when it is at the highest point on its trajectory. Where does the ball land? That is, what is the range of the ball?

Explanation / Answer

1) Height, h= 31m

Initial velocity in x direction ,Ux = 18m/s

Initial velocity in Y direction ( Taking upwards as positive) , Uy = 0

Final Velocity in X direction, Vx

Final Velocity in Y direction, Vy

Using Eqn of motion , s = ut + (1/2) at^2

-31= (Uy)t + (1/2)(-9.81)t^2

31 = (9.81/2)t^2

=> t = 2.514 sec ....................................... eqn 1

It takes 2.514 seconds to land

Using eqn of motion, s = ut + (1/2)at^2 , in x direction

D = (Ux)t + (1/2)at^2, .................. as a = 0 bocause there is no force in x direction

=> D= 18 X 2.514

=> D = 45.15 m

Vy = - gt

Vy = -9.81 X 2.514

Vy = -24.66 m/s

Magnitude of velocity = ((Vx)^2 + (Vy)^2)^.5

= ((18)^2 + (-24.66)^2)^.5

= 30.53 m/s

Direction = tan_inverse( Vy/Vx)

= -53.87 deg

2) Intial velocity, u = 28 m/s

Angle , A= 23 deg

Maximum height is given by , h_max = ( u sinB)^2/(2g)

h_max = ( 28 sin23)^2/(2*9.81)

= 6.1 m

Speed at highest point = u cosA .................. as there is no force in X axis to change the velocity

= 28*cos23

= 25.77 m/s

Time taken to go up and return = 2(2*h_max/g)^0.5 = 2.23 sec

Range = u*cosA*T

= 25.77*2.23

= 57.47 m/s

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