Question

A ball dropped from a wall of height h travels a distance of 50 m in last two seconds before landing. What is the height of the wall from which the ball was dropped?

Answer 1

Given :

• Distance traveled by ball in last 2 seconds = 50 m

To find :

• Height of the wall (h)

Solution :

• Let 't' be the time taken by the ball to travel the height h.
• The ball travels 50 m in the last 2 seconds.
• Hence, the ball would have traveled a distance of (h-50) m in the time of (t-2) seconds.
• Using the equation of motion,

        $s = ut + \frac{1}{2} gt^{2}$  

• The ball is dropped, this means the initial velocity will be 0.

      ∴  $h = 0 + \frac{1}{2} g t^{2}$  

      ∴  $t = \sqrt{\frac{2h}{g} }$     ....(1)

• Now, for time (t-2),

        $(h-50)=0+\frac{1}{2}g(t-2)^{2}$  

     ∴  $2h-100=g(t^{2} -4t+4 )$

• Substituting value of 't' from equation (1),

     ∴  $2h - 100 = g(\frac{2h}{g}-4\sqrt{\frac{2h}{g} }+4)$

     ∴  $2h-100=2h-4\sqrt{2gh} +4g$

     ∴  $4\sqrt{2gh} = 100 +4g$

     ∴  $\sqrt{2gh} = 25 +g$

• Taking value of g as 9.8 m/s²

        $19.6h = 34.8^{2}$

     ∴ h = 61.78 m

Answer :    

The height of the wall from which the ball was dropped is 61.78 m .