A projectile is launched from the ground, such that it hits a target on the ground which is $90 \mathrm{~m}$…
A projectile is launched from the ground, such that it hits a target on the ground which is $90 \mathrm{~m}$ away. The minimum velocity of projectile to hit the target is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
10. $\mathrm{ms}^{-1}$
$16 \mathrm{~ms}^{-1}$
$60 \mathrm{~ms}^{-1}$
$30 \mathrm{~ms}^{-1}$
Solution
Since, launched projectile hit the target on the ground $90 \mathrm{~m}$ away hence, range, $R=90 \mathrm{~m}$.
For minimum velocity $\left(u_{\min }\right)$ of projectile, range should be maximum. i.e. $\theta=45^{\circ}$
$R=\frac{u^2 \sin 2 \theta}{g}$
$\begin{array}{ll}\Rightarrow & 90=\frac{u_{\min }^2 \sin 2 \times 45^{\circ}}{10} \Rightarrow 900=u_{\min }^2 \\ \Rightarrow & u_{\min }=\sqrt{900}=30 \mathrm{~m} / \mathrm{s}\end{array}$