11.16m/s
The motion has both vertical and horizontal components. Using one of the equations of motion as follows;
h = ut + [tex]\frac{1}{2}[/tex] 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 + ([tex]\frac{1}{2}[/tex] 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 + ([tex]\frac{1}{2}[/tex] x 0 x t²)
12.30 = ut -------------------------------(ii)
Now, let's take the vertical component. Using the same equation;
h = ut + ([tex]\frac{1}{2}[/tex] 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 + ([tex]\frac{1}{2}[/tex] 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 = [tex]\sqrt{1.214}[/tex]
=> 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
