1.

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

Answer»

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%.


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