A particle is projected at an angle of $θ$ from the horizontal with the kinetic energy $T$ . What is the kinetic energy of the particle at the highest point?

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Solution

At initial point
T = $\frac{1}{2} m V^{2}$

At highest point

$T_{o} = \frac{1}{2} m (V c o s θ)^{2}$

= $\frac{1}{2} m V^{2} c o s^{2} θ$

$T_{0} = T c o s^{2} θ$

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A particle is projected at an angle of θ from the horizontal with the kinetic energy T. What is the kinetic energy of the particle at the highest point?

At initial point T = 12mV2At highest pointTo=12m(Vcosθ)2 = 12mV2cos2θ T0=Tcos2θ

A particle is projected at an angle of $θ$ from the horizontal with the kinetic energy $T$ . What is the kinetic energy of the particle at the highest point?

At initial point
T = $\frac{1}{2} m V^{2}$

At highest point

$T_{o} = \frac{1}{2} m (V c o s θ)^{2}$

= $\frac{1}{2} m V^{2} c o s^{2} θ$

$T_{0} = T c o s^{2} θ$