Search any question & find its solution
Question:
Answered & Verified by Expert
A body is projected vertically upwards at time $t=0$ and it is seen at a height $H$ at time $t_{1}$ and $t_{2}$ second during its flight. The maximum height attained is ( $g$ is acceleration due to gravity)
Options:
Solution:
1295 Upvotes
Verified Answer
The correct answer is:
$\frac{g\left(t_{1}+t_{2}\right)^{2}}{4}$
Let $t^{\prime}$ be the time taken by the body to fall from point $C$ to $B$.
Then $t_{1}+2 t^{\prime}=t_{2} \Rightarrow t^{\prime}=\left(\frac{t_{2}-t_{1}}{2}\right)$...(i)
Total time taken to reach point $C$
$$
\begin{aligned}
T &=t_{1}+t^{\prime}=t_{1}+\frac{t_{2}-t_{1}}{2} \\
&=\frac{2 t_{1}+t_{2}-t_{1}}{2}=\left(\frac{t_{1}+t_{2}}{2}\right)
\end{aligned}
$$
Maximum height attained
$$
\begin{array}{c}
H_{\max }=\frac{1}{2} g(T)^{2}=\frac{1}{2} g\left(\frac{t_{1}+t_{2}}{2}\right)^{2} \\
=\frac{1}{2} g \cdot \frac{\left(t_{1}+t_{2}\right)^{2}}{4} \\
\text { Or, } H_{\max }=\frac{1}{8} g \cdot\left(t_{1}+t_{2}\right)^{2} m
\end{array}
$$
Then $t_{1}+2 t^{\prime}=t_{2} \Rightarrow t^{\prime}=\left(\frac{t_{2}-t_{1}}{2}\right)$...(i)
Total time taken to reach point $C$
$$
\begin{aligned}
T &=t_{1}+t^{\prime}=t_{1}+\frac{t_{2}-t_{1}}{2} \\
&=\frac{2 t_{1}+t_{2}-t_{1}}{2}=\left(\frac{t_{1}+t_{2}}{2}\right)
\end{aligned}
$$
Maximum height attained
$$
\begin{array}{c}
H_{\max }=\frac{1}{2} g(T)^{2}=\frac{1}{2} g\left(\frac{t_{1}+t_{2}}{2}\right)^{2} \\
=\frac{1}{2} g \cdot \frac{\left(t_{1}+t_{2}\right)^{2}}{4} \\
\text { Or, } H_{\max }=\frac{1}{8} g \cdot\left(t_{1}+t_{2}\right)^{2} m
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.