Question

A projectile leaves the ground at an angle 60 to the horizontal.its kinetic energy is E . neglecting air resistance,find in terms of E it's kinetic energy at the highest point of the motion

Answer 1

Explanations & Calculations

• At the heighest point the velocity of the projectile is only the horizontal component.
• It should be known that the horizontal component of the velocity of a projectile remains unchaged thus it is $\small v\cos\theta=v \cos 60=\large \frac{v}{2}$
• Therefore,kinetic energy it has at the highest point is
• $\qquad\qquad \begin{aligned} \small E_{up}&=\small\frac{m(v\cos\theta)^2}{2}\\ &=\small \frac{mv^2}{8}\cdots(1) \end{aligned}$

• At the start its energy was,

$\qquad\qquad \begin{aligned} \small E&=\small\frac{mv^2}{2}\cdots(2) \end{aligned}$

• By (1)/(2),

$\qquad\qquad \begin{aligned} \small \frac{E_{up}}{E} &=\small\frac{1}{4}\\ \small E_{up}&=\small \frac{E}{4} \end{aligned}$