Question

A particle is projected vertically upwards from a point O on the ground. It takes a time $t_1$ to reach a point A at height h above the ground, it continues to move and takes a time $t_2$ to reach the ground, the maximum height attained by the particle will be?

• A. $\frac{g}{2}(t_1+t_2)$
• B. $\frac{g}{2}(t_1+t_2{)}^2$
• C. $\frac{g}{4}(t_1+t_2{)}^2$
• D. $\frac{g}{8}(t_1+t_2{)}^2$

Answer 1

Answer: (A)

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

Time taken for the particle to reach highest point is$\frac{t_1+t_2}{2}$

$(v=u-gt)$

initial velocity$=u=g=\left(\frac{t_1+t_2}{2}\right)$

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

$h=ut1-\frac{1}{2}g{t}_1^2$

$=g\left(\frac{t_1+t_2}{2}\right)t1-\frac{1}{2}g{t}_1^2$

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