Question

A particle projected vertically upwards reaches the maximum height H in T seconds. The height of the particle any time t will be

• A. $g(t-T{)}^2$
• B. $\frac{1}{2}g(t-T)$
• C. $\frac{1}{2}gt(2T-t)$
• D. $g(t-T)$

Answer 1

The correct option is C $\frac{1}{2}gt(2T-t)$

Given: The particle reaches a maximum height "H" in time 'T'.

Use equation of motion $v=u+aT$;

Take the upward direction sign as "-ve" and downward as "+ve";

$v=u-gT$

$0=u-gT$

$u=gT.$.....Eq:01

The height or the distance travelled (s) at time (t) will be;

$s=ut+\frac{1}{2}a{t}^2$

Put the value of 'u' from Eq:01;

$s=gT\left(t\right)+\frac{1}{2}(-g)t^2$

$s=gtT-\frac{1}{2}g{t}^2$

$s=\frac{1}{2}gt(2T-t)$

The required height of the particle at time (t) will be $\frac{1}{2}gt(2T-t)$

Option (C) is correct.