Question

A projectile is fired at 30° with momentum p.Neglecting air friction,the change in kinetic energy when it returns to the ground will be
a)zero
b)60%
c)30%
d)100%

Answer 1

Answer:

Given:

Projectile is fired at 30°.

To find:

Change in Kinetic Energy upon reaching the ground.

Calculation:

Let velocity of Projection be v

So, the initial Velocity can be divided into perpendicular components :

$v_{x} = v \cos( \theta)$

and

$v_{y} = v \sin( \theta)$

Now considering a Projectile to be consisting of 2 simultaneously occuring linear motions. , we can say that :

$v_{x} \: (reaching \: ground) = v \cos( \theta)$

and

$v_{y} \: (reaching \: ground) = - v \sin( \theta)$

Negative sign has been considered due to action of gravity downwards.

Now, initial Kinetic Energy:

$\boxed{ke1 = \frac{1}{2} m {v}^{2}}$

Final total velocity :

$|v_{final}| = \sqrt{ { {v}^{2} \cos( \theta) }^{2} + {v}^{2} { \sin( \theta) }^{2} } \\ = > |v_{final}| = v$

So final Kinetic Energy:

$\boxed{ke2 = \frac{1}{2} m {v}^{2}}$

So change in Kinetic Energy = 0%.