Question

B.
A particle is projected from ground at an angle OF 30° with horizontal. If kinetic energy at the ground is K, then potential energy at the highest point is
(1)k
(2) 4K
3)K/4
4)K/2​

Answer 1

The potential energy at the highest point will be $\dfrac{K}{4}$.

Explanation:

• Suppose, the particle is projected with the speed 'v' and its kinetic energy is given as 'K' at point A. (Refer the figure)

• We assume B is the highest point. We can understand that the vertical component of velocity become zero at highest point, only horizontal component works at B.

Now, from the conservation of energy, we can write as;

• (kinetic energy + potential energy)$_A$= (Kinetic energy + potential energy)$_B$

• $(K + 0) = (\dfrac{1}{2}mv^2)cos^2\theta + (PE)_B$

$K = Kcos^2\theta + (PE)_B$

$PE_B = K - K cos^2\theta$

$(PE)_B = K (1 - cos^2\theta)$

$(PE)_B = Ksin^2\theta$

$(PE)_B = K sin^2(30^o)$

$(PE)_B = \dfrac{K}{4}$

• This is the potential energy at the highest point.