A projectile is fired at 30• with momentum p, neglecting friction, c

1.	A projectile is fired at 30• with momentum p, neglecting friction, change in K. E, when it returns back to ground will be???
Answer» Given : PROJECTION angle of the PROJECTILE = 30° Momentum with which it is projected = p To FIND : Assuming no friction change in Kinetic energy when it RETURNS back to the ground = ? Solution : Since there is no friction , so we can apply Law of Energy CONSERVATION, i.e. Change in potential energy will be equal to change in Kinetic energy, so : $\Delta U = \Delta K E$ ∴U = mgh ⇒ΔU =mgΔh Since the projectile is back on the ground so Δh= 0 Hence ΔU = 0 Now since ΔU = 0 , so ΔKE should also be 0 . So, when the particle returns back to the ground , its change in K.E. will be 0 .

A projectile is fired at 30• with momentum p, neglecting friction, change in K. E, when it returns back to ground will be???

Answer»

Given :

PROJECTION angle of the PROJECTILE = 30°

Momentum with which it is projected = p

To FIND :

Assuming no friction change in Kinetic energy when it RETURNS back to the ground = ?

Solution :

Since there is no friction , so we can apply Law of Energy CONSERVATION, i.e. Change in potential energy will be equal to change in Kinetic energy, so :

$\Delta U = \Delta K E$

∴U = mgh

⇒ΔU =mgΔh

Since the projectile is back on the ground so Δh= 0

Hence ΔU = 0

Now since ΔU = 0 , so ΔKE should also be 0 .

So, when the particle returns back to the ground , its change in K.E. will be 0 .

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