CBSE Class 9 Answered
a stone is dropped from the top of a tower 100m high and instantly a second stone is projected vertically upward from the bottom with a velocity of 35/√2 metre per second.Find when and where the two stones meet ?
Asked by sheokanddimple | 16 Jan, 2023, 06:00: PM
Expert Answer
Let t seconds be the time duration for both the stones meet from starting time.
Distance h1 travelled by stone dropped from tower, h1 = (1/2) g t2
where g is acceleration due to gravity .
Distance h2 travelled by stone thrown vertically upwards from bottom,
h2 = ( u t ) - (1/2) g t2
where u = 35/√2 m/s is initial speed
h1 + h2 = (35/√2) t - (1/2) gt2 + (1/2) g t2 = 100 m
(35/√2) t = 100 m
t = ( 100 × √2 ) / 35 = 4.04 s
h1 = (1/2) × 9.8 × 4.04 × 4.04 ≈ 80 m
h2 = ( 100 - 80 ) m = 20 m
Hence stones will meet at a distance 20 m from bottom of tower after 4.04 s from initial starting time.
Answered by Thiyagarajan K | 16 Jan, 2023, 07:52: PM
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