Question

A projectile is fired from a cannon at a speed of 301 m/s and at an angle of 3 degrees. How long does it take the projectile to reach the highest point? How far does it go horizontally?

Answer 1

1. Time, required to reach the highest point is (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/):

$$t = \dfrac{v_0\sin\theta}{g}$$

where $v_0 = 301m/s$ is the initial speed, $\theta = 3\degree$ is the launch angle, and $g = 9.81m/s^2$ is the gravitational acceleration. Thus, obtain:

$$t = \dfrac{301\cdot \sin3\degree}{9.81} \approx 1.6\space s$$

2. The horizontal distacne is:

$$L = \dfrac{v_0^2\sin(2\theta)}{g} = \dfrac{301^2\cdot \sin(2\cdot 3\degree)}{9.81} \approx 965.4m$$

Answer. 1) 1.6 s, 2) 965.4 m.