Question

A body 'A' is projected upwards with a velocity of 98 m/s and a second body 'B' is projected upwards with the same initial velocity but after 4 seconds. Both the bodies will meet after

a) 6 seconds
b) 8 seconds
c) 10 seconds
d) 12 seconds

Please answer ASAP with explanation

Answer 1

Answer:  

Both the bodies will meet after 12 seconds.

Solution:

As the initial velocity at which the bodies A and B projected are same and also the position pf meeting will also be same so $\mathrm{H}_{\mathrm{A}}=\mathrm{H}_{\mathrm{B}}$

The height covered by the projecting body is  

$H=u t-\frac{1}{2} g t^{2}$

The time taken to reach the point H for object A is t

The time taken to reach the point H for object B is (t-4) as the object B is thrown after 4s of throwing A. So,

$H_{A}=H_{B}$

$u t-\frac{1}{2} g t^{2}=u(t-4)-\frac{1}{2} g(t-4)^{2}$

$u t-\frac{1}{2} g t^{2}=u t-4 u-\frac{1}{2} g\left(t^{2}-8 t+16\right)$

$-\frac{1}{2} g t^{2}=-4 u-\frac{1}{2} g t^{2}+4 g t-8 g$

$4 u-4 g t+8 g=0$

As u = 98 m/s , the above equation will become as  

$(4 \times 98)+(8 \times 9.8)=(4 \times 9.8) t$

$392+78.4=39.2 t$

$470.4=39.2 t$

$t=\frac{470.4}{39.2}=12 s$

Thus, it will take 12 s by object B to join object A.

Answer 2

Explanation:

Hope it helps you please follow me