A stone is thrown with speed 50m/s in vertical upward direction under uniform acceleration due to gravity. distance covered by the stone in last second and second last second of its upward motion are ..........
Answer is 5m, 15m.
but i couldn't understand. Please explain stepwise.
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Answer:
Given
initial velocity (u) = 50m/s
final velocity (v) =0
Acceleration (a) = -10 m/s^2
time (t) =?
we. know that
v=u+at => 0=50-10t => t = 5s
last second is 5th(n) second
distance covered in 5th second = u + a/2 (2n -1)
=50-10/2 (2(5)-1)
= 50-45
= 5 m
distance covered in 4th(n) second = u + a/2(2n-1)
=50-10/2(2(4)-1)
=50-35
=15 m
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