TIFR CSE 2020 | Part A | Question: 14

Question

A ball is thrown directly upwards from the ground at a speed of $10\: ms^{-1}$, on a planet where the gravitational acceleration is $10\: ms^{-2}$. Consider the following statements:

1. The ball reaches the ground  exactly $2$ seconds after it is thrown up
2. The ball travels a total distance of $10$ metres before it reaches the ground
3. The velocity of the ball when it hits the ground is $10\: ms^{-1}$

What can you say now?

• A. Only Statement $1$ is correct
• B. Only Statement $2$ is correct
• C. Only Statement $3$ is correct
• D. None of the Statements $1,2$ or $3$ is correct
• E. All of the Statements $1,2$ and $3$ are correct

Answer 1

According to the equations of motion for a particle in a straight line under uniform acceleration :

$1)\; v = u + at$

$2)\; s = ut + \frac{1}{2}at^2$

$3)\; v^2 = u^2 + 2as$

$1):$ When ball is thrown directly upwards from the ground

In this case, $u=10\; m/sec,$ $a=g= -10 \;m/s^2$(because I assumed displacement $s$ as positive in upward direction) and $v=0$ at maximum height.

So, from $(1)$, $0= 10 - 10t \Rightarrow t = 1 \;sec$

from $(2)$, $s= 10*1 - \frac{1}{2}*10*1^2 \Rightarrow s = 5 \;m$

$2):$ When ball goes downwards after reaching the maximum height

In this case, $u=0\; m/sec,$ $a=g= +10 \;m/s^2$(because I assumed displacement $s$ as negative in downward direction)

So, from $(3)$, $v^2= 0 + 2*10*5 \Rightarrow v = 10 \;m/sec$

from $(1)$, $10= 0 + 10*t \Rightarrow t = 1 \;sec$

So, from $(1)$ and $(2)$

$1)$ Total time in journey : $1+1 = 2\; sec$

$2)$ Total displacement in journey : $5+5 =\; 10\; m$

$3)$ Velocity of the ball when it hits the ground : $10\;m/sec$