Question

A boy on a cliff 49m high drops a stone. One second later, he throws a second after the first. They both hit the ground at the same time. With what speed did he throw the second stone?
(a) 12.1 m/s
(b) 10 m/s
(c) 21.1 m/s
(d) 15.1 m/s

Answer 1

Hint:
Use the 2nd equation of motion to find the time(t) taken by stones to reach inn their respective directions. Logic is since acceleration is constant and motion is 1D equations of motions can be applicable to calculate distance and time relation.

Formula used:
2nd Equation of motion:
$y = ut + \dfrac{1}{2}a{t^2}$ …… (1)
where,
y is the distance covered by a particle in vertical direction.
u is initial velocity.
a is constant acceleration
‘t’ is time taken to cover ‘y’ distance.

Complete step by step solution:
Given:
1. Height of cliff (y)$ = 49m$
2. Initial velocity for 1st stone =0m/s

To find: The initial velocity of the 2nd stone.

Step 1 of 3:
Let t be the time taken for the first stone to hit the ground. Construct the 2nd equation of motion for the first stone (using eq 1):
$49 = \dfrac{1}{2}(9.8){t^2}$
Find t by rearranging eq (2): …… (2)
\[
  t = \sqrt {\dfrac{{98}}{{9.8}}} \\
  t = \sqrt {10} \\
 \]

Step 2 of 3:
Let the time taken by the second stone to hit the ground be $t'$. Now, time taken for second stone to hit the ground will be:
$
  t' = t - 1 \\
  t' = \sqrt {10} - 1 \\
  t' = 2.16s \\
 $

Step 3 of 3:
Let the initial velocity of the second stone be u. Construct the 2nd equation of motion for the second stone (using eq 1):
\[
  49 = ut' + \dfrac{1}{2}(9.8)t{'^2} \\
  49 = u(2.16) + 4.9{(2.16)^2} \\
 \]
Rearrange to get u:
$
  u = \dfrac{{49 - 4.9{{(2.16)}^2}}}{{2.16}} \\
  u = \dfrac{{26.13}}{{2.16}} \\
  u = 12.1m/s \\
 $

Note:
One should always remember equations of motions are only applicable in 1D motion with constant acceleration. Higher dimensional motions like projectiles can be divided into two 1D motions, along each axis to apply equations of motions.