A ball is released from a height 'h' reaches the ground in time 'T'. Where will it be from the ground at time 3T/4 ?
(a) 9h/16
(b) 7h/16
(c) 3h/4
(d) 27h/64
*No guessing please, give detailed explanation.
Answers
Answered by
67
Since,the ball is released from rest, its initial velocity is zero. When released from height h it reaches the ground in time T.
If we apply the second equation of motion,
s = ut+at²/2
h = gT²/2
T² = 2h/g
Now, we apply the same equation to calculate the height covered in time 3T/4.
s = ut+at²/2
s = g× (3T/4)²/2
= (9g/32)T²
= (9g/32)× 2h/g
= 9h/16 from the point of release.
Distance from the ground = h - 9h/16
= 7h/16
Hence, option b) is correct.
If we apply the second equation of motion,
s = ut+at²/2
h = gT²/2
T² = 2h/g
Now, we apply the same equation to calculate the height covered in time 3T/4.
s = ut+at²/2
s = g× (3T/4)²/2
= (9g/32)T²
= (9g/32)× 2h/g
= 9h/16 from the point of release.
Distance from the ground = h - 9h/16
= 7h/16
Hence, option b) is correct.
Answered by
10
Answer:
Answer is option b
Explanation:
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