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VITEEE 2009 Paper

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Question : 2 of 120

Marks: +1, -0

A body is projected vertically upwards at time t = 0 and it is seen at a height H at time

t_{1}\text{ and }t_{2}

second during its flight. The maximum height attained is (g is acceleration due to gravity)

$\frac{g (t_{2}-t_{1})^{2}}{8}$
$\frac{g (t_{1}+t_{2})^{2}}{4}$
$\frac{g (t_{1}+t_{2})^{2}}{8}$
$\frac{g (t_{2}-t_{1})^{2}}{4}$

Solution:

Let t' be the time taken by the body to fall from point C to B.

Then

t_{1}+2t'=t_{2}

⇒

t'=\left(\frac{t_{2}-t_{1}}{2}\right)

...(i)

Total time taken to reach point C

T=t_{1}+t'=t_{1}+\frac{t_{2}-t_{1}}{2}=\frac{2t_{1}+t_{2}-t_{1}}{2}=\left(\frac{t_{1}+t_{2}}{2}\right)

Maximum height attained

H_{max}=\frac{1}{2}g (T)^{2}=\frac{1}{2}g \left(\frac{t_{1}+t_{2}}{2}\right)^{2}

=\frac{1}{2}g \cdot \frac{(t_{1}+t_{2})^{2}}{4}

Or,

H_{max}=\frac{1}{8}g \cdot (t_{1}+t_{2})^{2}\text{m}

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