Question

A cannonball of mass 5.99 kg is shot from a cannon at an angle of 50.21° relative to the horizontal and with an initial speed of 52.61 m/s. As the cannonball reaches the highest point of its trajectory, what is the gain in its potential energy relative to the point from which it was shot

Answer 1

Given:

m=5.99kg

α=50.21°

u=52.61 m/sec

To Find:

The gain Potential Energy at the highest point of trajectory.

Explanation:

u=Initial speed

α=Projection angle

m=mass

The formula for maximum height in a projectile is given by:

$H=\frac{u^{2} sin ^{2} \alpha}{2g}$

The gravitational potential energy is given by :

mgh, here h is the maximum height of the projectile we just defined.

$P.E.=mgh$

Putting h=H:

$P.E.=mg\frac{u^{2} sin ^{2} \alpha}{2g}$

$P.E.=\frac{mu^{2} sin ^{2} \alpha}{2}$

Put in the given values from question:

P.E.=4894.43 Joules

Final Answer: Potential Energy=4894.93 Joules