Question

A particle is projected vertically upwards from a point A on the ground. it take t1 time to reach a point B but still continue to move up.if it takes further t2 time to reach the ground from point B then height of point B from ground is?
answer is 1/2gt1t2 ​

Answer 1

Answer:

$\frac{1}{2}gt_1t_2$

Explanation:

1) Let the  speed of particle at position A be 'u' and at B be 'v' .

It took $t_1$ time to reach B from A.

Then , using Newton's Equation of motion,

$v=u-gt_1$

2) It took $t_2$ time to reach from B to A after covering maximum height .

We know, when particle reaches the ground its speed will be 'u'. (downward)

Again,

$-u=v-gt_2$

3) Using above two equations to get ,

$u=\frac{1}{2}g(t_1+t_2)$

4) We  know, it took $t_1$ time to move from A to B ,

Using Newton's Equation of motion,

$S=ut_1-\frac{1}{2}gt_1^2\\ \\ =\frac{1}{2}g(t_1+t_2)t_1-\frac{1}{2}gt_1^2\\ \\=\frac{1}{2}gt_1^2+\frac{1}{2}gt_1t_2-\frac{1}{2}gt_1^2\\ \\=\frac{1}{2}gt_1t_2$

Hence, Height of point 'B' from ground is

$\boxed{\frac{1}{2}gt_1t_2}$

Answer 2

Explanation:

Total time of particle =t1+t 2

to find initial value

Applying 2nd equation of motion

o=u(t1+t2 )− g/2 (t1+t2)^2

⇒ g/2 (t1+t2 )=u [∵ height at t=t1 +t2 is zero]

At t=t1 , height attained is l

So,

h=ut1 − g/2t1 ^2

h= g/2(t1+t2 )t1 −/2g t1^2

h=g/2t1 [t1+t2 −t]

h= g/2t1t2

The height of point B(h) is 2gt1t2

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