Question.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms–2, with what velocity will it strike the ground? After what time will it strike the ground ?
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms–2, with what velocity will it strike the ground? After what time will it strike the ground ?
Solution:
Given, initial velocity of ball, $\mathrm{u}=0$
Final velocity of ball, $\mathrm{v}=?$
Distance through which the ball falls, $\mathrm{s}=20 \mathrm{~m}$
Acceleration $\mathrm{a}=10 \mathrm{~ms}^{-2}$
Time of fall, $\mathrm{t}=$ ?
We know
$v^{2}-u^{2}=2 a s$
or $\mathrm{v}^{2}-0=2 \times 10 \times 20=400$ or $\mathrm{v}=20 \mathrm{~ms}^{-1}$
Now using $\mathrm{v}=\mathrm{u}+$ at we have
$20=0+10 \times t$ or $t=2 s$
Given, initial velocity of ball, $\mathrm{u}=0$
Final velocity of ball, $\mathrm{v}=?$
Distance through which the ball falls, $\mathrm{s}=20 \mathrm{~m}$
Acceleration $\mathrm{a}=10 \mathrm{~ms}^{-2}$
Time of fall, $\mathrm{t}=$ ?
We know
$v^{2}-u^{2}=2 a s$
or $\mathrm{v}^{2}-0=2 \times 10 \times 20=400$ or $\mathrm{v}=20 \mathrm{~ms}^{-1}$
Now using $\mathrm{v}=\mathrm{u}+$ at we have
$20=0+10 \times t$ or $t=2 s$