Explore BrainMass

# Physics: Trajectory of a Projectile

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

** Please see the attached file for the complete problem description **

A projectile (m=2kg) is shot from a cannon at an initial angle of &#952;=30° with an initial speed of vo=40m/s. Assume the ball is shot from the origin.

https://brainmass.com/physics/conservation-of-momentum/physics-trajectory-projectile-364140

#### Solution Preview

** Please see the attached file for the complete solution response **

a) Maximum height reached by the projectile is given by: H = u^2 sin^2 θ/2g

Substituting u = 40 m/s, θ = 30^O we get: H = 40^2 sin^2 30^O /2x9.8 = 20.41 m

Maximum height reached = 20.41 m

To determine the time taken to reach the maximum height, we note that the vertical component of the initial velocity u_y = usin30^O = 40sin30^O = 20 m/s

As the projectile moves up, it is subjected to a vertical retardation of g = - 9.8 m/s^2

We now determine the time taken by the projectile to cover a distance of 20.41 m, starting with a vertical velocity of 20 m/s and decelerating at - 9.8 m/s^2. Applying the kinematic equation y = u_y t + ½ at^2 we get: 20.41 = 20t - ½ x 9.8 x t^2

4.9t^2 - 20t + 20.41 = 0
__________________
Solving for t: t = [- (- 20) + √(- 20)^2 - 4(4.9)(20.41)]/2(4.9)
________
t = [+ 20 + √400 - 400]/9.8 = 20/9.8 = 2.04 sec

Projectile takes 2.04 ...

#### Solution Summary

This solution provides detailed explanations of how to determine a projectile's trajectory based on it's angle and initial speed. A number of other measures are determined in this tutorial.

\$2.49