Question

a body a is projected upwards with B velocity of 98 m/s. the second body B is projected upwards with the same initial velocity but after 4 sec. both the bodies will meet after

Answer 1

Answer:

Both the bodies have been thrown with same initial velocity

So, h1 ​= h2​

Time of flight form for 1st body = t second

and for 2nd body = (t- 4) second  (because it is projected 4sec later)

SO, h1 ​= h2​

ut−1/2gt^2 = 4(t−4)−1/2​g(T−4)^2

By solving the equation

t = 12second

Explanation: