Question

How much high above the ground a boy can throw the ball if he is able to throw the same ball up to maximum horizontal distance of 50m?

Answer 1

The horizontal range is maximum (R-max) when the angle of projection is 45 degree.
R-max = (u^2)/g
So, (u^2)/g = 50 m ...(i)

For vertical motion,
Maximum Height (H)
= (u^2)/2g
= (u^2)/g × (1/2) ...(just rearranged the above equation)
= 50 × (1/2)
= 25 m

Thus, the maximum height till which the boy can throw the ball is 25 m.

Answer 2

Answer:

The maximum height that the boy can throw the ball is 25 m.

Explanation:

Given the maximum horizontal distance a ball has thrown is R = 50m.

The horizontal distance or the range of projectile is given by

$R=\dfrac{u^{2}sin2\theta }{g}$

The horizontal distance covered by the projectile is maximum when the angle of projection is 45°

$\therefore R=\dfrac{u^{2}sin(2\times 45^o) }{g} =\dfrac{u^{2}sin90^o}{g} =\dfrac{u^{2}}{g}$

And hence, the maximum range is

$R_{max}=\dfrac{u^{2}}{g}$

Substituting the value of the maximum range given,

$50=\dfrac{u^{2}}{g}$

The vertical distance or the height of the projectile is given by

$H=\dfrac{u^{2}sin^2\theta }{2g}$

For maximum height to be reached, $\theta=90^o$ and hence,

$H_{max}=\dfrac{u^{2}sin^2(90) }{2g}=\dfrac{u^{2}}{2g}$

        $=\dfrac{u^{2}}{g}\times\dfrac{1}{2}=50\times\dfrac{1}{2}=25m$

Therefore, the maximum height the boy can throw the ball is 25 m.