Question:
If a pebble is thrown vertically upwards from the top of a tower with velocity \(5\text{ m/s}\). It strikes the ground after \(3\) seconds. With what velocity does the pebble strike the ground? \((g=10\text{ ms}^{-2})\)
If a pebble is thrown vertically upwards from the top of a tower with velocity \(5\text{ m/s}\). It strikes the ground after \(3\) seconds. With what velocity does the pebble strike the ground? \((g=10\text{ ms}^{-2})\)
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Choose a sign convention carefully. If upward is positive, then acceleration due to gravity is negative.
Updated On: May 7, 2026
- \(10\text{ m/s}\)
- \(20\text{ m/s}\)
- \(25\text{ m/s}\)
- \(30\text{ m/s}\)
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The Correct Option is C
Solution and Explanation
The pebble is thrown vertically upward with initial velocity:
\[
u=5\text{ m/s}.
\]
Take upward direction as positive.
Acceleration due to gravity acts downward, so:
\[
a=-g=-10\text{ m/s}^2.
\]
Time taken is:
\[
t=3\text{ s}.
\]
We need the velocity after \(3\) seconds.
Use the first equation of motion:
\[
v=u+at.
\]
Substitute the values:
\[
v=5+(-10)(3).
\]
\[
v=5-30.
\]
\[
v=-25\text{ m/s}.
\]
The negative sign shows that the pebble is moving downward when it strikes the ground.
The question asks the velocity with which it strikes the ground, so we take the magnitude:
\[
|v|=25\text{ m/s}.
\]
Hence, the pebble strikes the ground with velocity:
\[
25\text{ m/s}.
\]
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