1
Question
A coin is thrown in a vertically upward direction with a velocity of 5 m s−1. If the acceleration of the coin during its motion is 10 m s−2 in the downward direction, what will be the height attained by the coin and how much time will it take to reach there respectively?
A coin is thrown in a vertically upward direction with a velocity of 5 m s−1. If the acceleration of the coin during its motion is 10 m s−2 in the downward direction, what will be the height attained by the coin and how much time will it take to reach there respectively?
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Solution
The correct option is B 1.25 m, 0.5 s
Here,
Initial velocity, u=5 m s−1, and
acceleration, a=−10 m s−2
At the highest point,
final velocity, v=0
Using, v2=u2+2as,
0=52+[2×(−10)×s]
⇒20s=25
⇒ Distance, s=2520=1.25 m
Hence, the height attained by the coin is 1.25 m.
Let, the time taken be t.
using v=u+at,
0=5−10t
⇒t=510=0.5 s
Thus, time taken by the coin to reach there is 0.5 s.
Here,
Initial velocity, u=5 m s−1, and
acceleration, a=−10 m s−2
At the highest point,
final velocity, v=0
Using, v2=u2+2as,
0=52+[2×(−10)×s]
⇒20s=25
⇒ Distance, s=2520=1.25 m
Hence, the height attained by the coin is 1.25 m.
Let, the time taken be t.
using v=u+at,
0=5−10t
⇒t=510=0.5 s
Thus, time taken by the coin to reach there is 0.5 s.
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