Question

A projectile is thrown with an angle of with x-axis
such that it has a kinetic energy of 100J at the point
of projection. What will be the potential energy of
projectile at the highest point?
A:
100J
B:
75J
С:
50J
D:
25J​

Answer 1

The projectile has only kinetic energy at the point of projection.

$\longrightarrow\sf{K_1=\dfrac{1}{2}\,mu^2}$

$\longrightarrow\sf{U_1=0}$

So we can say that the total energy of the projectile is equal to its kinetic energy at the point of projection.

$\longrightarrow\sf{E=K_1=\dfrac{1}{2}\,mu^2}$

We know total energy is conserved throughout a projectile motion.

At highest point the projectile has horizontal velocity only, so its kinetic energy will be,

$\longrightarrow\sf{K_2=\dfrac{1}{2}\,mu^2\cos^2\theta}$

The potential energy of the projectile at the highest point, therefore, will be,

$\longrightarrow\sf{U_2=E-K_2}$

$\longrightarrow\sf{U_2=\dfrac{1}{2}\,mu^2-\dfrac{1}{2}\,mu^2\cos^2\theta}$

$\longrightarrow\sf{U_2=\dfrac{1}{2}\,mu^2\left[1-\cos^2\theta\right]}$

$\longrightarrow\sf{U_2=\dfrac{1}{2}\,mu^2\sin^2\theta}$

$\longrightarrow\sf{U_2=E\sin^2\theta}$

Here, $\sf{E=K_1=100\ J.}$ So,

$\longrightarrow\sf{U_2=100\sin^2\theta}$

For $\sf{\theta=90^o,}$

$\longrightarrow\sf{U_2=100\sin^290^o}$

$\longrightarrow\sf{U_2=100\times1}$

$\longrightarrow\sf{U_2=100\ J}$

For $\sf{\theta=60^o,}$

$\longrightarrow\sf{U_2=100\sin^260^o}$

$\longrightarrow\sf{U_2=100\times\dfrac{3}{4}}$

$\longrightarrow\sf{U_2=75\ J}$

For $\sf{\theta=45^o,}$

$\longrightarrow\sf{U_2=100\sin^245^o}$

$\longrightarrow\sf{U_2=100\times\dfrac{1}{2}}$

$\longrightarrow\sf{U_2=50\ J}$

For $\sf{\theta=30^o,}$

$\longrightarrow\sf{U_2=100\sin^230^o}$

$\longrightarrow\sf{U_2=100\times\dfrac{1}{4}}$

$\longrightarrow\sf{U_2=25\ J}$