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The horizontal and vertical displacements of a projectile at time $t$ are $x=36 t$ and $y=48 t-4.9 t^2$, respectively. Initial velocity of the projectile in $\mathrm{m} / \mathrm{s}$ is
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Verified Answer
The correct answer is:
$60$
$$
x=36 t
$$
$$
y=48 t-4.9 t^2
$$
Comparing with $x=u \cos \theta t$
$$
y=u \sin \theta t-\frac{1}{2} g t^2
$$
where, $u=$ initial velocity
$$
\begin{aligned}
& u \cos \theta=36, \\
& u \sin \theta=48
\end{aligned}
$$
$$
\begin{aligned}
(u \cos \theta)^2+(u \sin \theta)^2 & =36^2+48^2 \\
u^2\left(\cos ^2 \theta+\sin ^2 \theta\right) & =3600 \\
u^2 & =3600 \\
u & =60 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
x=36 t
$$
$$
y=48 t-4.9 t^2
$$
Comparing with $x=u \cos \theta t$
$$
y=u \sin \theta t-\frac{1}{2} g t^2
$$
where, $u=$ initial velocity
$$
\begin{aligned}
& u \cos \theta=36, \\
& u \sin \theta=48
\end{aligned}
$$
$$
\begin{aligned}
(u \cos \theta)^2+(u \sin \theta)^2 & =36^2+48^2 \\
u^2\left(\cos ^2 \theta+\sin ^2 \theta\right) & =3600 \\
u^2 & =3600 \\
u & =60 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
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