Question

A ball is thrown horizontally from the top of tower 100mhight,withspeed10m/s.after 2sec of fall gravity disappears.then total time taken by ball to reach the ground is(g=10m/s)

Answer 1

Answer:

Total time taken by the ball to reach the ground = 6 seconds

Time taken by the ball to reach the ground after the gravity disappears = 4 seconds

EXPLAINATION:

Given,

The height of the tower = 100 m

Initial horizontal speed = 10 m/s

Ball is thrown horizontally from the top of the tower therefore its initial vertical speed will be zero

Thus using the second equation of motion

The vertical distance h covered in 2 seconds

\boxed{h=\frac{1}{2}gt^2}

h=

2

1

gt

2

or, h=\frac{1}{2}\times 10\times 2^2=20h=

2

1

×10×2

2

=20 m

Therefore the height left = 100 - 20 = 80 m

The vertical velocity after 2 seconds

using the first equation of motion

v=0+gtv=0+gt

\implies v=10\times2=20⟹v=10×2=20 m/s

Since the gravity disappears after 2 seconds

therefore there will be no acceleration in the vertical direction

time taken to cover 80 m distance = distance/velocity

= 80/20

= 4 second

Answer 2

Explanation:

use the time of flight ... 2 u sin theta by g..