Question
A ball thrown vertically upwards with a speed of 20 ms−1 from the top of a tower reaches the earth in 8 s. Find the height of the tower. Take g=10 ms−2
Answer: Option D
:
D
Given:
Initial velocity, u=20 ms−1
Time taken, t=8 s
Let the height of the tower be h.
Assume the upwards direction to be positive.
The magnitude of the displacement is the height of the tower. Final position (ground) is in downwards (or negative) direction with respect to the initial position (top of tower).
∴s=−h
Acceleration is in downwards direction.
∴a=−g=−10 ms−2
From the second equation of motion,
s=ut+12at2,
⇒−h=ut−12gt2
⇒−h=20×8−(12×10×82)
⇒h=160 m.
therefore, height of tower = 160 m
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:
D
Given:
Initial velocity, u=20 ms−1
Time taken, t=8 s
Let the height of the tower be h.
Assume the upwards direction to be positive.
The magnitude of the displacement is the height of the tower. Final position (ground) is in downwards (or negative) direction with respect to the initial position (top of tower).
∴s=−h
Acceleration is in downwards direction.
∴a=−g=−10 ms−2
From the second equation of motion,
s=ut+12at2,
⇒−h=ut−12gt2
⇒−h=20×8−(12×10×82)
⇒h=160 m.
therefore, height of tower = 160 m
Was this answer helpful ?
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