A stone thrown horizontally from a height of 6.22 m hits the ground at a distance of 11.90 m. Calculate the initial speed of the stone. Neglect air resistance.

10.625m/s

Explanation:

Using one of the equations of motion as follows;

h =  u t + a t²       -------------------(i)

Where;

h = horizontal/vertical distance covered

u = initial velocity of the object(stone)

t = time taken to cover the distance

a = acceleration of the object.

Taking the horizontal motion of the stone;

h = horizontal distance of the stone  = 11.90m

a = acceleration = 0 (the horizontal motion is at constant velocity, therefore, acceleration is 0).

Substitute these values into equation(i) to give;

11.9 = u t + 0

11.9 = ut   ------------------------------(ii)

Taking the vertical motion of the stone,

h = height from which the stone was thrown = 6.22m

a = acceleration due to gravity = g = +10m/s² (since the stone moves downwards in the direction of gravity)

u = 0 (since the ball was thrown horizontally, the vertical component of the velocity, u, is 0)

Substitute these values into equation (i) to give;

6.22 = 0 x t + x 10 x t²      

6.22 =  5 x t²       --------------------(iii)

Solve for t;

6.22 =  5t²

5t² = 6.22

t² = 6.22 / 5

t² = 1.244

t =

t = 1.12s

Since it takes the same time for the stone to hit the ground as it will take it to move the horizontal distance of 11.9, therefore we can substitute the time t = 1.12s into equation (ii) as follows;

=> 11.9 = ut

=> 11.9 = u (1.12)

=> u = 11.9 / 1.12

=> u = 10.625m/s

Therefore, the initial speed of the stone is 10.625m/s

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