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A body is projected at an angle other than $90^{\circ}$ with the horizontal with same velocity. If the time of ascent of the body is ls, then the maximum height it can reach is (Take, $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
5 m
Given, angle of projection, $\theta=90^{\circ}$ and time of ascent, $t=1$ s.
Since, time of ascent of a projectile body is half of the time of height, $T=2 \times 1=2 \mathrm{~s}$
Equation of motion,
$$
\mathrm{s}=u t+\frac{1}{2} g t^2
$$
where, $s=0$ for complete flight.
and gravitational acceleration, $g=10 \mathrm{~m} / \mathrm{s}^2$
Putting the given values in above equation, we get
$$
\begin{aligned}
& 0=u \times 2-\frac{10}{2} \times 2^2 \\
& u=10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Now, the maximum height,
$$
h=u t-\frac{10}{2} t^2
$$
where, $t=$ ls (time of ascent)
$$
\begin{aligned}
& h=10 \times 1-5 \times 1 \\
& h=5 \mathrm{~m}
\end{aligned}
$$
Hence, the correct option is (a).
Since, time of ascent of a projectile body is half of the time of height, $T=2 \times 1=2 \mathrm{~s}$
Equation of motion,
$$
\mathrm{s}=u t+\frac{1}{2} g t^2
$$
where, $s=0$ for complete flight.
and gravitational acceleration, $g=10 \mathrm{~m} / \mathrm{s}^2$
Putting the given values in above equation, we get
$$
\begin{aligned}
& 0=u \times 2-\frac{10}{2} \times 2^2 \\
& u=10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Now, the maximum height,
$$
h=u t-\frac{10}{2} t^2
$$
where, $t=$ ls (time of ascent)
$$
\begin{aligned}
& h=10 \times 1-5 \times 1 \\
& h=5 \mathrm{~m}
\end{aligned}
$$
Hence, the correct option is (a).
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