Question

2 point
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/s2)​

Answer 1

Answer: 19.6 m/s

Let the initial speed be = u m/s

After travelling for 5 seconds, the particle takes 9 seconds to come back.

--› The particle was in air for 4 seconds more after covering a distance of *h* m .

--› The particle took 2 seconds more after attaining height *h* to attain it's maximum height.

Hence, total time taken to reach max height,

*H* = 5 + 2 = 7 sec.

Applying equation of motion,

$v = u + at$

At max Height *H*, v = 0 m/s

t = 7 sec

$→0 = u + ( - g)t \\ →u = gt \\ →u = 9.8 \times 7 \\ →u = 68.6$

Therefore, at height *h* speed of the particle will be :

$v_{h} = 68.6 - (9.8 \times 5) \\ →v_{h} = 68.6 - 49.0 \\ →v_{h} = 19.6 \: m {s}^{ - 1}$

Hope this helps.