Solveeit Logo

Question

Question: A ball rolled off the edge of a horizontal table with a speed of \(4m{{s}^{-1}}\) , if it reaches th...

A ball rolled off the edge of a horizontal table with a speed of 4ms14m{{s}^{-1}} , if it reaches the ground after 0.3 sec. Then the speed with which it strikes the ground is (g=10ms2)\left( g=10m{{s}^{-2}} \right)
A. 45ms14\sqrt{5}m{{s}^{-1}}
B. 5ms15m{{s}^{-1}}
C. 52ms15\sqrt{2}m{{s}^{-1}}
D. 3ms13m{{s}^{-1}}

Explanation

Solution

In order to solve this problem choose an appropriate equation of motion or more than one equations from the equations of motion as per the given data and then try to solve for the unknown variables required.

Complete step by step answer:
The equations of motion we are having are:
v=u+atv=u+at
s=ut+12at2s=ut+\dfrac{1}{2}a{{t}^{2}}
v2u2=2as{{v}^{2}}-{{u}^{2}}=2as
Where v=final velocityv=\text{final velocity}
u=initial velocityu=\text{initial velocity}
a=accelerationa=\text{acceleration}
t=timet=\text{time}
s=distance travelleds=\text{distance travelled}
The data given is
Initial speed of the ball in horizontal direction is u=4ms1u=4m{{s}^{-1}}
Initial speed of the ball in vertical direction is u=0ms1u=0m{{s}^{-1}}
Time taken by the ball the reach the ground is t=0.3sect=0.3\sec
Acceleration due to gravity g=10ms2g=10m{{s}^{-2}}
Since the ball is falling down we need to consider the initial vertical velocity not the initial horizontal velocity
Here the ball is travelling due to gravity hence the acceleration of the ball is taken as the acceleration due to gravity i.e., gg with the proper sign conventions that is ggis taken negative (ve)\left( -ve \right) for the ball travelling away from the earth (moving upwards)\left( \text{moving upwards} \right) and is taken positive(+ve)\left( +ve \right) for the ball travelling towards the earth (moving downwards)\left( \text{moving downwards} \right)
According to the data we are having the appropriate equation from the above equations of motion we are having is
v=u+atv=u+at
Here a=ga=g
Since the ball is moving towards the earth gg is taken as positive
v=u+gt\Rightarrow v=u+gt
.Substitute the above given values in this equation
v=u+gtv=u+gt
v=0+10×(0.3)\Rightarrow v=0+10\times \left( 0.3 \right)
v=3ms1\therefore v=3m{{s}^{-1}}
Therefore the required velocity with which the ball touches the ground is v=3ms1v=3m{{s}^{-1}}
Hence option (D)(D) is correct.

Note:
One should take care while dealing with the one dimensional and two dimensional system of motion and the forces should be balanced accordingly, that is by resolving the components of motion in x and y directions separately and then they should be balanced by using the equations of motion. Also care should be taken while handling the sign conventions in order to avoid calculation errors.