Question

A ball is thrown vertically upwards with a velocity of $20m{s^{ - 1}}$ from the top of a multi storey building. The height of the point where the ball is thrown $25m$ from the ground. How long will it be before the ball hits the ground? Take $g = 10m{s^{ - 2}}$.
A. t = 5s
B. t = 10s
C. t = 15s
D. t = 20s

Answer 1

Hint: We need to use the equations for motion for solving this problem. In those equations, we also need to replace acceleration by acceleration due to gravity because the ball is moving under the effect of gravity here.

Formula used:
Equations of motion are as follows:
$v = u + at$
$s = ut + \dfrac{1}{2}a{t^2}$
${v^2} = {u^2} + 2as$

Complete step-by-step answer:
We are given that a ball is thrown upwards with initial velocity given as
$u = 20m{s^{ - 1}}$
We know that the velocity at maximum height will be zero, so $v = 0$
Here the initial velocity is in upward direction and acceleration in downward so the acceleration will come with negative sign, i.e., retarded motion.
$a = - g = - 10m{s^{ - 2}}$
Now we can calculate the time required $\left( t \right)$ to reach the maximum height, $\left( H \right)$
For calculating the time $\left( t \right)$, we first need to apply the first equation of motion which can be written as follows.
$\Rightarrow v = u + at$
$\Rightarrow 0 = 20 + \left( { - 10} \right) \times t$
$\Rightarrow 10t = 20$
$\Rightarrow t = \dfrac{{20}}{{10}}$
$\Rightarrow t = 2s$
Now we can calculate the maximum height $\left( H \right)$by applying the second equation of motion,
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow H = 20 \times 2 + \dfrac{1}{2} \times \left( { - 10} \right){\left( 2 \right)^2}$
$\Rightarrow H = 40 - 20$
$\Rightarrow H = 20m$
Hence, the total height is given as,

$$= 20m + 25m \ = 45m \\\$$

Now we can calculate the time required to reach the ground $\left( {t'} \right)$, from maximum height using the second equation of motion in the following way.
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2} \\\ \Rightarrow 45 = 0 + \dfrac{1}{2} \times \left( {10} \right) \times {\left( {t'} \right)^2} \\\ \Rightarrow 5{{t'}^2} = 45 \\\ \Rightarrow {{t'}^2} = \dfrac{{45}}{5} \\\ \Rightarrow {{t'}^2} = 9 \\\ \Rightarrow t' = 3s \\\$
Hence, the total time is given as
$\Rightarrow t + t' = 2 + 3 = 5s$
Hence, the correct option is A, i.e., 5s

Additional Information:
Students should proceed in the following manner for solving problems related to the equation of motion:
1. Read the question carefully to identify the given quantities and note them.
2. Identify the equation to use and note them.
3. Ensure that all the values are in the same system of measurement and put them in identified equations.
4. Calculate the answer carefully and check the final units.

Note: Students should memorize the three equations of motion and understand the physical significance of these equations. Students need to know how to apply the suitable equation of motion on the basis of given data. Students should ensure that all the quantities are in the same system of measurement i.e., S.I., M.K.S., C.G.S. etc.