Question

A particle is thrown upwards. It attains a height (h) after 5 seconds and again after 9 seconds comes back. What is the speed of the particle at a height h?

Answer 1

Given that,

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

$v=u-gt$

$0=u-9.8\times 0.5$

$u=4.9m/s$

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