Question

A particle is projected vertically upwards from a point A on the ground. It takes t, time to reach a point B but
it still continues to move up. If it takes further t, time to reach the ground from point B then height of point B
from the ground is :
(A) 1/2g(t1+t2) ^2
(B) gt1t2
(C) 1/8g(t1+t2) ^2
(D) 1/2gt1t2​

Answer 1

D) 1/2g$t_{1} t_{2}$

Explanation:

Given that,

Time = t

so,

Total time = $t_{1} + t_{2}$

Time required to reach the optimum height = ( $t_{1} + t_{2}$)/2

Now, Let 'u' be the particle's initial velocity and 'h' be the height of point B

By applying equation I of motion,

0 = u + (−g) [ $t_{1} + t_{2}$]

⇒ u = [g( $t_{1} + t_{2}$)]/2

Now we know that, at t =  $t_{1}$,

Using IInd equation of motion, we get

h = u$t_{1}$ - 1/2g$t_{1^{2} }$

= [g( $t_{1} + t_{2}$)  $t_{1}$]/2 - g$t_{1^{2} }$/2

∵ h = 1/2g$t_{1} t_{2}$

Thus, option D is the correct answer.

Learn more: Motion

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