Question

2 point
1. A particle is thrown
upwards. It attains a height (h)
after 5 seconds and again after
9scomes back. What is the speed
of the particle at a height h?(take
g=9.8m/s)​

Answer 1

Velocity v=um/s

Time taken to reach at maximum height t

1 = 5s

Now,

Total time taken to reach same point t

2 = 9s

Particle to reach its maximum height will be half of total time =4.5s

Now,

the particle returns to the same point in time t=9−5

=4s

Now,

the time at height is

t=4.5−4

t=0.5s

So,

the velocity at height h

v=u−gt

u−9.8×0.5

u=4.9m/s

Hence, the speed of the particle at height is 4.9 m/s.

Hope this helped you✌️✌️