Question

a coastguard ship locates a pirate ship at a distance of 560m if fired a cannon ball with a initial speed 82m/s at what angle from horizontal the ball must be fired so that it hits the pirate ship​

Answer 1

Given:

A coastguard ship locates a pirate ship at a distance of 560m. It fired a cannon ball with a initial speed 82m/s.

To find:

Angle of projection for hitting the pirate ship ?

Calculation:

This is an example of GROUND-GROUND projectile. Let angle of projection be $\theta$.

$\sf \therefore \: R = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\sf \implies\: 560 = \dfrac{ {(82)}^{2} \sin(2 \theta) }{10}$

$\sf \implies\: 560 = \dfrac{ 6724 \times \sin(2 \theta) }{10}$

$\sf \implies\: \sin(2 \theta) = \dfrac{5600}{6724}$

$\sf \implies\: \sin(2 \theta) = 0.83$

$\sf \implies\: 2 \theta = {56.09}^{ \circ}$

$\sf \implies\: \theta = {28.04}^{ \circ}$

So, angle of projection is 28.04°.

Answer 2

I have given the clue for which the question should be proceeded