Question

A stone thrown horizontally from a height of 6.07 m hits the ground at a distance of 12.30 m. Calculate the initial speed of the stone. Neglect air resistance.

Answer 1

Answer:

11.16m/s

Explanation:

The motion has both vertical and horizontal components. Using one of the equations of motion as follows;

h = ut + $\frac{1}{2}$ x a x t²

Where;

h = horizontal or vertical displacement

u = initial velocity in the horizontal/vertical motion

a = acceleration in the horizontal/vertical motion

t = time taken for the motion

Now, let's take the horizontal component first. Using the same equation;

h = ut + ($\frac{1}{2}$ x a x t²)           -----------------(i)

Where;

h = horizontal displacement = 12.30m

a = 0      [acceleration in the horizontal is zero as velocity is constant in that direction]

Substitute these values into equation (i) as follows;

12.30 = ut  + ($\frac{1}{2}$ x 0 x t²)

12.30 = ut       -------------------------------(ii)

Now, let's take the vertical component. Using the same equation;

h = ut + ($\frac{1}{2}$ x a x t²)           -----------------(iii)

Where;

h = vertical displacement = 6.07m

a = acceleration due to gravity = +10m/s²      [acceleration in the vertical is the acceleration due to gravity which can be positive or negative depending on the direction of motion. +ve if motion is upwards, and -ve if motion is downwards. In our case, the motion is downwards, hence +ve gravity]

u = 0        [there is no velocity in the vertical motion since the stone is thrown horizontally]

Substitute these values into equation (iii) as follows;

=> 6.07 = ut + ($\frac{1}{2}$ x 10 x t²)    

=> 6.07 = ut + 5t²     ---------------------(iv)

Substitute this into equation (iv) as follows;

=> 6.07 = 5t²

Solve for t;

=> 5t² = 6.07

=> t² = 6.07 / 5

=> t² = 1.214

=> t = $\sqrt{1.214}$

=> t = 1.102s

Substitute this value of t = 1.102 into equation (ii) as follows;

12.30 = u(1.102)

Solve for u;

u = 12.30 / 1.102

u = 11.16m/s

Therefore, the initial speed of the stone is 11.16m/s