Question

If the time taken by ball A to fall back on ground is 4 sec, the maximum height attained by ball A from the ground is:
$A.10m$
$B.15m$
$C.20m$
$D.$Insufficient information

Answer 1

Hint: We have to use the concept of projectile motion in terms of maximum height attained and time taken to attain maximum height. After finding this we will use a suitable equation of motion to get the correct solution.
Formula used:
We will use the following formulae to solve this question:-
$T=\dfrac{2u\sin \theta }{g}$and${{v}^{2}}-{{u}^{2}}=2as$.

Complete step by step solution:
We consider the above case as projectile motion and thus we have the following parameters with us:-
Total time of flight,$T=4s$.
Acceleration due to gravity,$g=10m/{{s}^{2}}$.
Angle of projection,$\theta $ for maximum vertical height is taken as ${{90}^{o}}$.
Suppose the ball is thrown with initial velocity,$u$.
Now using $T=\dfrac{2u\sin \theta }{g}$ to find initial velocity for the above given case we get
$4=\dfrac{2u\sin {{90}^{o}}}{10}$………………… $(i)$
Putting the value $\sin {{90}^{o}}=1$in equation $(i)$ we get
$4\times 10=2u$
$u=20m/s$………………….. $(ii)$
We know that at the maximum height its final velocity,$v=0m/s$.
Now using third equation of motion${{v}^{2}}-{{u}^{2}}=2as$,this equation is modified into
${{v}^{2}}-{{u}^{2}}=2gh$because we are dealing with the height and acceleration due to gravity.
We have ${{v}^{2}}-{{u}^{2}}=2gh$………………. $(iii)$
Putting the values of different parameters in $(iii)$ we get
$0-{{(20)}^{2}}=2\times -10\times h$
We have used $-g$ because the motion is against the direction of acceleration due to gravity.
Solving further we get
$-400=-20h$
Cancelling minus signs both sides and solving further we get
$h=20m$.
Therefore maximum height attained is $20m$.

Hence, option $(C)$ is the correct one.

Note:
We have to take care of the parameters of the projectile. We should not get confused between the terms time of flight and time of ascent. There are many similar relationships in projectile and we should not be confused among them. Sign of $g$should be taken according to the direction of motion.