Question

A body is projected up with a speed 'u' and the time taken is 'T' to reach the maximum height 'H'. Pick out the correct statements

Statement 1) It reaches (H/2) in (T/2) seconds.

Statement 2) It acquires velocity (u/2) in (T/2) seconds.

Statement 3) It's velocity is (u/2) at (H/2).

Statement 4) Same velocity at 2T

Pleaseeeeeee answer with explanation ASAP

Answer 1

I think 2nd is the correct option.
Hope i m correct:-)

Answer 2

Answer:

Concept:

The method or technique of projecting the points of a spatial object onto a plane, curved surface, or a line to create a graph or figure. When one vector is resolved into two constituent vectors, one parallel to the second vector and the other perpendicular to the second vector, the vector projection is created. The vector projection is the parallel vector.

Given:

The time it takes for a body to achieve the highest height 'H' is 'T', and the speed is 'u'.

Statement [1] It reaches (H/2) in (T/2) S.

Statement [2] It acquires velocity (u/2) in (T/2) S.

Statement [[3]] Its Velocity Is (u/2) At (H/2).....

Statement [4] Same velocity at 2T

Find:

Pick out the correct statement for the given question

Answer:

The correct option is statement [2] It acquires velocity (u/2) in (T/2) seconds

For a body thrown upward

we know that,

$S=ut-\frac{1}{2} gt^2$

For maximum height $v=0$

$v=u-gt=0$

$T=u/g$

So,

$H=(s)_{max} =u^{2} /g-u^{2} -2g$

                   $=\frac{u^{2} }{2g}$

$v=u-gt$

$At$ $t=\frac{T}{2}$

$V=u-g(\frac{u}{2g} )$

   $=\frac{u}{2}$

$v^2=u^2-2gs$

$so,$

$at$ $S=\frac{H}{2}$

$v^2=u^2-2g \frac{u^{2} }{2g}$

    $=\frac{u^{2} }{2g}$

$V=\frac{u}{\sqrt{2} }$

The projectile can reach a maximum height of h equal to one-half of H, the triangle's altitude H – ½H so h = H/2, which is the desired outcome. The object's maximum height is determined by the highest vertical spot along its flight path.

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