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Answer 3) 40 m/s.
A particle is thrown upwards with a velocity u and it raises to a height h. Let it take n seconds in its total flight. So it takes n/2 seconds upwards and n/2 sec. on its way down. Distance covered in n/2 th second is 35 m, before hitting the ground.
Distance S_t covered during the m th second is given by the formula:
S_m = U + (m - 1/2) a,
here U = 0, a = g = 10 m/s^2. m = n/2.
35 = 0 + (n/2 - 1/2) 10
n = 8 sec.
n/2 = 4 sec. the particle takes to reach maximum height.
So apply v = u + a t for the flight upwards.
0 = u - 10 * 4
u = 40 m/s.
A particle is thrown upwards with a velocity u and it raises to a height h. Let it take n seconds in its total flight. So it takes n/2 seconds upwards and n/2 sec. on its way down. Distance covered in n/2 th second is 35 m, before hitting the ground.
Distance S_t covered during the m th second is given by the formula:
S_m = U + (m - 1/2) a,
here U = 0, a = g = 10 m/s^2. m = n/2.
35 = 0 + (n/2 - 1/2) 10
n = 8 sec.
n/2 = 4 sec. the particle takes to reach maximum height.
So apply v = u + a t for the flight upwards.
0 = u - 10 * 4
u = 40 m/s.
kvnmurty:
:-) :-)
thakuu so much sir
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Explanation:
Option 2 is correct answer !!
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