Question

A ball is thrown vertically upward with a velocity of 20 m per second the distance covered by the ball during the first second in the downward journey is ​

Answer 1

Answer:

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

t=2s

s=ut+0.5at

2

=20(2)+0.5(−10)(2)

2

=20m

Thus, total distance for maximum height is 45 m

s=ut+0.5at

2

45=0+0.5(10)(t

′

)

2

t

′

=3s

Total time= 3+2= 5s

Answer 2

Answer:

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

t=2s

s=ut+0.5at

2

=20(2)+0.5(−10)(2)

2

=20m

Thus, total distance for maximum height is 45 m

s=ut+0.5at

2

45=0+0.5(10)(t)2t

=3s

Total time= 3+2= 5s