# Answer on Mechanics | Relativity Question for kshitij sharma

Question #15097

a ball which is thrown Vertically up words reaches the roof of a house 100 m high.at the moment this ball is thrown vertically up ward ,another ball is dropped from rest vertically downwards from the roof of the house.at which time, height the ball will pass each other.

Expert's answer

H = V0 * t - g * t^2 / 2

V0 = g * t, so t = V0 / g:

H = V0^2 / g - g *

V0^2 / (2 * g^2) = V0^2 / (2 * g)

V0 = sqrt(2 * g * H)

V0 = sqrt(2 * 9.81

* 100) = 44.30 m/s

For the first ball: Hx = V0 * tx - g * tx^2 /

2:

For the second one: Hx = H - g * tx^2 / 2

V0 * tx - g * tx^2 / 2 = H -

g * tx^2 / 2

V0 * tx = H

tx = H / V0

tx = 100 / 44.30 = 2.26 s

And

the height will be: Hx = H - g * tx^2 / 2, Hx = 100 - 9.81 *

2.26^2 / 2 =

74.95 m

V0 = g * t, so t = V0 / g:

H = V0^2 / g - g *

V0^2 / (2 * g^2) = V0^2 / (2 * g)

V0 = sqrt(2 * g * H)

V0 = sqrt(2 * 9.81

* 100) = 44.30 m/s

For the first ball: Hx = V0 * tx - g * tx^2 /

2:

For the second one: Hx = H - g * tx^2 / 2

V0 * tx - g * tx^2 / 2 = H -

g * tx^2 / 2

V0 * tx = H

tx = H / V0

tx = 100 / 44.30 = 2.26 s

And

the height will be: Hx = H - g * tx^2 / 2, Hx = 100 - 9.81 *

2.26^2 / 2 =

74.95 m

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