1.

A ball is thrown vertically upwards, with a speed of 10m/s from the top of a tower 200m high and another is thrown vertically downwards, with the same speed simultaneously. What is the time difference between them in reaching ground?

Answer»

HEYA MATE,

HERE IS UR ANSWER.

Thanks for asking this question.

The total time taken by Ball-1 from t-0 to the moment it REACHES ground be t-1 and Ball-2 be t-2

We need (t-1) - (t-2)

t-1

Upward phase, u = 10m/s a= -10m/s^2 v=0, time = v-u/a = -10/-10 = 1 second

Height attained during intial phase = ut + 1/2gt^2 = 10*1 + 1/2(-10)(1)^2 = 5m

Downward Phase, u = 0 height = 205 m,

==> 205m = 0 + 1/2(10)(t^2)

t= SQRT(41)

t-1 = 6.4031+1 = 7.4031

t-2

Height traveled = 200 m, u = 10m/s

==> 200 m = 10t + 1/2(10)(t^2)

5t^2+10t-200 = 0

t^2 + 2t - 40 = 0

t= [-2(+-)Sqrt(4–(4)(1)(-40))]/2 = [-1(+-)(1/2(Sqrt(164))] = -1 (+-) 12.8062/2 = 5.4031

Difference = 7.4031–5.4031 = 2 Seconds.

This big solution is offered as other solutions have already been given.

I HOPE IT HELPS U.


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