If it had been thrown with twice the speed in the same direction, it would have hit the ground in If it had been thrown with twice the speed in the same direction, it would have hit the ground in
3.20 s.
1.60 s.
0.800 s.
0.400 s.
0.200 s.

The ball will reach the ground in 0.8s

Option C

Explanation:

Given:

- Takes t = 0.8 s for ball to reach ground when thrown horizontal from top of a building.

Find:

If it had been thrown with twice the speed in the same direction, it would have hit the ground in how many second.

Solution:

- We know that the amount of time taken to hit the ground is determined by the vertical distance i.e height at which it is thrown. The displacement of ball from top is given by:

                                    S_y = S_o + V_i,y*t + 0.5*g*t^2

- We know that the S_o = height of the building.

 We also know that the ball os thrown horizontally; hence, y-component of initial velocity is zero. V_y,i = 0

                                    0 = h + 0 + 0.5*g*t^2

- Hence, the time taken t is:

                                    t = sqrt ( 2h / g)

- The time taken to reach the ground is independent of the initial speed. Hence, the ball will reach the ground in 0.8s .

1/0.14nm = 10^10/1.4 a lot