Question

A boy can throw a stone to a maximum height of 50 m. To what maximum range can he throw this stone and to what height so that the maximum range is maintained?

Answer 1

Hint: To solve this problem you must know the third equation of motion and formula for maximum range in projectile motion. To solve this question we apply a direct formula as given for projectile motion. To use it we can put values by observing the data given in the question.
Formula used:
Third equation of motion: ${v^2} = {u^2} + 2as$
Maximum range in projectile motion: ${R_{\max }} = \dfrac{{{u^2}}}{g}$
Greatest height for maximum range: ${h_{\max }} = \dfrac{{{u^2}}}{{2g}}{\sin ^2}\theta $
Where, $v = $ Final velocity of the ball
$u = $Initial velocity of the ball
$a = $Acceleration of the ball
$s = $ Displacement covered by the ball
$g = $ Acceleration due to gravity $ = 10m.{s^{ - 2}}$
$\theta = $ Angle of projection

Complete step-by-step answer:
It is given that the boy can throw a stone to a maximum height of 50 m.
To throw a stone to maximum height, the boy has to throw the stone vertically upwards.
Third equation of motion: ${v^2} = {u^2} + 2as$
Final velocity of the ball, $v = 0$
Acceleration of the ball = -Acceleration due to gravity = $ - g$
Displacement of the ball, $s = 50m$
Putting all these values to the third equation of motion, we get
$ \Rightarrow 0 = {u^2} - 2g \times 50$
$ \Rightarrow {u^2} = 100g$…………………. (1)
Maximum range, ${R_{\max }} = \dfrac{{{u^2}}}{g}$
Putting values in above equation, we get
${R_{\max }} = \dfrac{{100g}}{g} = 100m$
Now, maximum height for maximum range, ${h_{\max }} = \dfrac{{{u^2}}}{{2g}}{\sin ^2}\theta $
Putting values in above equation, we get
${h_{\max }} = \dfrac{{100g}}{{2g}}{\sin ^2}45^\circ $
$\theta = 45^\circ $ , for maximum range.
\[{h_{\max }} = \dfrac{{50}}{2}m = 25m\]
Hence, Maximum range of the ball is 100 m and greatest height for maximum range is 25 m.

Note: In this question we can solve it by using a direct formula or we can first draw a projectile diagram for given data then simply apply all forces to draw its path. We must keep in view we will use horizontal and vertical components of particles and we draw its horizontal and vertical movement after we can find restaurant movement.