Kinetic Energy and Parabolic Motion

1. Dec 19, 2011

1. The problem statement, all variables and given/known data
A shell of mass 50kg is fired at 60 degress to the horizontal with a speed of 200 m/s. Neglecting air resistance,calculate the kinetic energy of the shell at the highest point.

2. Relevant equations
K.E=.5(m)(v)^2
usin(θ)
ucos(θ)

3. The attempt at a solution
My question is shouldnt the K.E of the shell is 0? Since the vertical velocity is zero at the highest point. Should I use horizontal velocty instead? why is that so?

If i use horizontal velocity: 200cos60=100
K.E=(1/2)(50)(100)^2=2.5*10^5

2. Dec 19, 2011

Staff: Mentor

You need the total velocity to calculate the KE. But since the vertical component is zero at the highest point, all you need is the horizontal component.

Exactly. (Don't forget units.)

3. Dec 19, 2011

Staff: Mentor

KE is the energy of motion. The shell is still moving when it's at its zenith (it's moving horizontally).

You could calculate the KE in terms of components, vertical and horizontal, then add them like you add vector components to find the magnitude. But since you know that the vertical component is zero when the shell is at its maximum height, the calculation simplifies to being just the KE due to the horizontal component.

4. Dec 19, 2011

bjd40@hotmail.com

yes u have to use the horizontal component of velocity in this case.
remember the velocity is zero at the topmost point only when the projectile is fired vertically, i.e, with zero horizontal component of velocity. if it is fired with some angle other than 90 deg with the horizontal it has a non zero velocity at the top. that velocity = the initial horizontal component of the velocity. this remains unchanged through out the flight as the gravitational force is a vertical (downward) one and as such can not affect the horizontal component of velocity.