Question

The maximum distance to which a man can throw a ball by projecting it horizontally from a height h is h. The maximum distance to which he can throw it vertically up is
A. h
B. 2h
C. h/2
D. h/4

Answer 1

Hint: Ball is not thrown from the ground whereas it is thrown from a certain height and horizontally. The time taken to reach the ground can be calculated and within that time the horizontal distance travelled is already given. We can find out the velocity of projection from this.

Formula used:
$\eqalign{
  & s = ut + \dfrac{1}{2}a{t^2} \cr
  & {v^2} - {u^2} = 2as \cr} $

Complete step-by-step answer:
From a height ‘h’ a ball is thrown horizontally. That means it is given only with horizontal velocity but not any vertical velocity. Acceleration due to gravity ‘g’ acts along vertically downward and there is no acceleration along the horizontal direction.
We will calculate the time taken to reach the ground by using formula $s = ut + \dfrac{1}{2}a{t^2}$ where s=h and a=g and initial vertical component of velocity is zero i.e u=0.
$s = ut + \dfrac{1}{2}a{t^2}$
$\eqalign{
  & \Rightarrow h = \dfrac{1}{2}g{t^2} \cr
  & \Rightarrow t = \sqrt {\dfrac{{2h}}{g}} \cr} $
With in that time the horizontal distance covered is given as ‘h’ we will find out the horizontal velocity of projection from the formula $s = ut + \dfrac{1}{2}a{t^2}$ where acceleration along x direction is zero and $t = \sqrt {\dfrac{{2h}}{g}} $
$s = ut + \dfrac{1}{2}a{t^2}$
$\eqalign{
  & \Rightarrow h = u\sqrt {\dfrac{{2h}}{g}} \cr
  & \Rightarrow u = \sqrt {\dfrac{{gh}}{2}} \cr} $
If the ball is thrown with this velocity vertically upward then he distance travelled by ball will be calculated by formula ${v^2} - {u^2} = 2as$ where ‘v’ is the final velocity which will be zero in this case as we need to calculate the maximum height travelled. Acceleration is ‘-g’ as it is downward direction and $u = \sqrt {\dfrac{{gh}}{2}} $ by substituting all these values in ${v^2} - {u^2} = 2as$ we get
${v^2} - {u^2} = 2as$
$\eqalign{
  & \Rightarrow {0^2} - {u^2} = 2( - g)s \cr
  & \Rightarrow {u^2} = 2gs \cr
  & \Rightarrow s = \dfrac{{{u^2}}}{{2g}} \cr} $
$\eqalign{
  & \Rightarrow s = \dfrac{{{{\left( {\sqrt {\dfrac{{gh}}{2}} } \right)}^2}}}{{2g}} \cr
  & \Rightarrow s = \dfrac{h}{4} \cr} $
So the maximum distance which he can throw up is $\dfrac{h}{4}$

So, the correct answer is “Option D”.

Note: Initially we had found the velocity with which he had thrown the ball horizontally because it is the maximum velocity with which he can throw because it was given in the question that maximum horizontal range is attained when thrown with that velocity and he should throw with same velocity to get the maximum vertical range too.