Answer to Question #283432 in Mechanics | Relativity for Naba

Question

two possible angle to hit a target by a martyr shell fired with initial velocity of 98 metre per second are 15 degree and 75 degree calculate the range of projectile and the minimum time to required to hit the target

Answer 1

Explanations & calculations

• The above 2 angles are the results obtained upon solving the formula $\small R=\large\frac{v^2\sin2\theta}{g}$ for angles for a given velocity and a given range.
• Therefore, the range that would result with any of those angles is $\qquad\qquad \begin{aligned} \small R&=\small \frac{(98\,ms^{-1})^2 \sin2(15)}{9.8\,ms^{-2}}=\frac{98^2\sin2(75)}{9.8}\\ &=\small 490\,m \end{aligned}$
• Once this is found you can use $\bold{\color{blue}s=ut}$ for the shell's horizontal motion to calculate the time that it would take to reach the $\small 490\,m$ destination.
• Here $u$ is the horizontal resolution of the shell's initial velocity at any of the given angles.

• You can give it a try now, can't you?