Question
A coin is thrown in a vertically upward direction with a velocity of 5 ms−1. If the acceleration of the coin during its motion is 10 ms−2 in the downward direction, what will be the height attained by the coin and how much time will it take to reach there ?
Answer: Option D
:
D
Given, initial velocity, u=5ms−1, acceleration, a=−10ms−2.
Let 'v' be the final velocity and 's' be the height attained.
At the highest point of motion, v=0
Using third equation of motion, v2=u2+2as,
0=52+[2×(−10)×s]
⇒s=1.25m
Hence the height attained by the coin is 1.25m.
Let the time taken be t.
Using the first equationof motion,
v=u+at
0=5−10t
⇒t=0.5s
∴ Time taken by the coin is 0.5s.
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:
D
Given, initial velocity, u=5ms−1, acceleration, a=−10ms−2.
Let 'v' be the final velocity and 's' be the height attained.
At the highest point of motion, v=0
Using third equation of motion, v2=u2+2as,
0=52+[2×(−10)×s]
⇒s=1.25m
Hence the height attained by the coin is 1.25m.
Let the time taken be t.
Using the first equationof motion,
v=u+at
0=5−10t
⇒t=0.5s
∴ Time taken by the coin is 0.5s.
Was this answer helpful ?
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