Question

14. An object takes 3 s to reach the ground when thrown downwards from a point with speed 5 m/s. It takes
4 s when thrown upwards from that point with the same speed. How long does it take to reach the ground
when dropped from rest from this point? [Take g = 10 m/s2]​

Answer 1

Explanation:

Explanation :- case 1 :-when body moves upward direction.

initial velocity =u

final velocity =v=0

now, v=u+at

Here, v=0,a=−g [acceleration is downward direction]

0=u−gt⇒t=u/g

Maximum height = h

u

2

=2gh⇒h=u

2

/2g---(1)

Case 2 :- when body moves downward direction.[Falling from maximum height]

initial velocity, u=0

final velocity = v

use formula, s=ut+1/2at

2

here,s=−h,u=0 and a=−g

−h=0−1/2gt

′2

−u

2

/2g=−gt

′2

/2 [from equation(1)]

t

′

=u/g

Hence, total time =t+t

′

=u/g+u/g=2u/g

Answer 2

Answer:

Explanation:

Explanation :- case 1 :-when body moves upward direction.

initial velocity =u

final velocity =v=0

now, v=u+at

Here, v=0,a=−g [acceleration is downward direction]

0=u−gt⇒t=u/g

Maximum height = h

u

2

=2gh⇒h=u

2

/2g---(1)

Case 2 :- when body moves downward direction.[Falling from maximum height]

initial velocity, u=0

final velocity = v

use formula, s=ut+1/2at

2

here,s=−h,u=0 and a=−g

−h=0−1/2gt

′2

−u

2

/2g=−gt

′2

/2 [from equation(1)]

t

′

=u/g

Hence, total time =t+t

′

=u/g+u/g=2u/g