A body is projected vertically upwards. If $${t_1}$$ and $${t_2}$$ be the times at which it is at height $$h$$ above the projection while ascending and descending respectively, then $$h$$ is
A.
$$\frac{1}{2}g{t_1}{t_2}$$
B.
$$g{t_1}{t_2}$$
C.
$$2g{t_1}{t_2}$$
D.
$$2hg$$
Answer :
$$\frac{1}{2}g{t_1}{t_2}$$
Solution :
$$\eqalign{
& h = u{t_1} - \frac{1}{2}gt_1^2 \cr
& {\text{Also}}\,h = u{t_2} - \frac{1}{2}gt_2^2 \cr} $$
After simplify above equations, we get
$$h = \frac{1}{2}g{t_1}{t_2}.$$
Releted MCQ Question on Basic Physics >> Kinematics
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