Question:
If a pebble is thrown vertically upwards from the top of a tower with velocity \(5\,m/s\), it strikes the ground after \(3\) seconds. With what velocity the pebble strikes the ground? \((g=10\,m/s^2)\)
If a pebble is thrown vertically upwards from the top of a tower with velocity \(5\,m/s\), it strikes the ground after \(3\) seconds. With what velocity the pebble strikes the ground? \((g=10\,m/s^2)\)
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For vertical motion, carefully choose sign convention. If downward is positive, upward velocity is negative.
Updated On: May 7, 2026
- \(10\,m/s\)
- \(20\,m/s\)
- \(25\,m/s\)
- \(30\,m/s\)
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The Correct Option is C
Solution and Explanation
Concept:
For vertical motion under gravity, use: \[ v=u+gt \] when downward direction is taken as positive.
Step 1: The pebble is thrown vertically upwards. If downward direction is taken as positive, then initial velocity is: \[ u=-5\,m/s \]
Step 2: Acceleration due to gravity is: \[ g=10\,m/s^2 \]
Step 3: Time taken is: \[ t=3\,s \]
Step 4: Use: \[ v=u+gt \] \[ v=-5+10(3) \] \[ v=-5+30 \] \[ v=25\,m/s \]
Step 5: Therefore, the pebble strikes the ground with velocity: \[ \boxed{25\,m/s} \]
For vertical motion under gravity, use: \[ v=u+gt \] when downward direction is taken as positive.
Step 1: The pebble is thrown vertically upwards. If downward direction is taken as positive, then initial velocity is: \[ u=-5\,m/s \]
Step 2: Acceleration due to gravity is: \[ g=10\,m/s^2 \]
Step 3: Time taken is: \[ t=3\,s \]
Step 4: Use: \[ v=u+gt \] \[ v=-5+10(3) \] \[ v=-5+30 \] \[ v=25\,m/s \]
Step 5: Therefore, the pebble strikes the ground with velocity: \[ \boxed{25\,m/s} \]
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