Question

For a particle projected with speed of $20m{s}^{-1}$ from ground, the kinetic energy at top most point of trajectory is $\frac{1}{4}th$ of the initial kinetic energy. Then the angle of projection from horizontal is:

Answer 1

Step 1: Given data

Initial speed of the projectile, $u=20m{s}^{-1}$

The kinetic energy at the topmost point, $K{E}_{top}=\frac{1}{4}K{E}_i$

where $K{E}_i$ is the initial kinetic energy of the projectile

Step 2: Formula used

$KE=\frac{1}{2}m{v}^2$

where $m$ is mass and $v$ is velocity of the particle.

Step 3: Finding the kinetic energy at topmost point

We know, at the topmost point vertical component of velocity is zero.

Therefore, there is only horizontal component.

Horizontal component of velocity, $v_x=u\cos \theta$

Kinetic energy at topmost point,

$$\begin{gathered} K{E}_{top}=\frac{1}{2}m{v_x}^2 \\ \Rightarrow K{E}_{top}=\frac{1}{2}m{\left(u\cos \theta \right)}^2 \end{gathered}$$

Step 2: Finding the initial kinetic energy and comparing them

Initial kinetic energy, $K{E}_i=\frac{1}{2}m{u}^2$

Since,

$$\begin{gathered} K{E}_{top}=\frac{1}{4}K{E}_i \\ \Rightarrow \frac{1}{2}m{u}^2{\cos }^2\theta =\frac{1}{4}\times \frac{1}{2}m{u}^2 \\ \Rightarrow c{\mathrm{os}}^2\theta =\frac{1}{4} \\ \Rightarrow \cos \theta =\frac{1}{2} \\ \Rightarrow \theta =60° \end{gathered}$$

Therefore, the angle of projection is $60°$.