First slide

Projection Under uniform Acceleration

Question

A projectile is projected with kinetic energy E. If it has the maximum possible horizontal range, then its kinetic energy at the highest point will be

Easy

A

0.25 E
B

0.5 E
C

0.75 E
D

E

Solution

Here $θ = 45^{\circ}$ because range is maximum.
So, $V_{x} = Vcos 45^{\circ} = V / \sqrt{2}$
K.E. at highest Point
$= \frac{1}{2} m (v / \sqrt{2})^{2} = \frac{1}{4} {mv}^{2} = \frac{E}{2} = 0 \cdot 5 E$

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