Question

# A ball is thrown horizontally from the top of a building hits the ground 0.25 s...

A ball is thrown horizontally from the top of a building hits the ground 0.25 s later, Ignore air resistance. If it had been thrown with twice the speed in the same direction, it would have hit the ground in:

A) 0.40 s

B) 0.50 s

C) 1.00 s

D) 0.25 s

Given that ball is thrown horizontally from the top of building,

Suppose height of building is 'h', then when ball hits on the ground, then

Vertical displacement of ball = -h (-ve since in downward direction)

Using 2nd kinematic equation in vertical direction:

d = V0y*t + (1/2)*a*t^2

V0y = Initial vertical speed = 0 m/s, since ball was thrown horizontally

a = acceleration due to gravity = -g = -9.81 m/s^2

So,

-h = 0*t + (1/2)*(-g)*t^2

t = sqrt (2*h/g)

So we can see that time taken by ball to land on ground only depends on the height of building and not on the initial horizontal velocity, So if we double initial launching speed, then will be no effect.

So In this case too ball will land after 0.25 sec

Correct option is D.

Let me know if you've any query.

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