Answer to Question #249129 in Mechanics | Relativity for mhia

Question

A Cannon is fired with a muzzle velocity of 300 m/s, at an angle of 60 degrees. Find the

range, time of flight, maximum height attained by the projectile.

Answer 1

The range, time of flight and maximum height are given as follows respectively (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/)

$$R = \dfrac{u^2\sin2\theta}{g} \\ T = \dfrac{2u\sin\theta}{g}\\ h = \dfrac{(u\sin\theta)^2}{2g}$$

where $u = 300m/s$ is the initial speed, $\theta =60\degree$ is the launch angle, $g = 9.81m/s^2$ is the gravitational acceleration.

Substituting the values, obtain:

$$R = \dfrac{300^2\cdot \sin(2\cdot 60\degree)}{9.81}\approx 7945m\\ T = \dfrac{2\cdot 300\cdot \sin60\degree}{9.81}\approx 53.0s\\ h = \dfrac{(300\cdot \sin60\degree)^2}{2\cdot 9.81} \approx 3440m$$

Answer. 7945m, 53s, 3440m.