A particle is projected making an angle of 45° with horizontal having kinetic energy K. The kinetic energy at highest point will be

Question

A particle is projected making an angle of 45° with horizontal having kinetic energy K. The kinetic energy at highest point will be (A) (K/√ 2) (B) (K/2) (C) 2K (D) K

Tardigrade Admin · Accepted Answer

Kinetic energy of the ball =K and angle of projection (θ)=45°. Velocity of the ball at the highest point =v cos θ =v cos 45°=(v/√2). Therefore kinetic energy of the ball =(1/2) m ×((v/√2))2=(1/4) m v2=(K/2).

Q. A particle is projected making an angle of $4 5^{\circ}$ with horizontal having kinetic energy $K$ . The kinetic energy at highest point will be

$\frac{K}{2}$

$\frac{K}{2}$

$2 K$

$K$

Kinetic energy of the ball $= K$
and angle of projection $(θ) = 4 5^{\circ}$ .
Velocity of the ball at the highest point
$= v cos θ = v cos 4 5^{\circ} = \frac{v}{2}$ .
Therefore kinetic energy of the ball
$= \frac{1}{2} m \times (\frac{v}{2})^{2} = \frac{1}{4} m v^{2} = \frac{K}{2}$ .

Q. A particle is projected making an angle of 45∘ with horizontal having kinetic energy K. The kinetic energy at highest point will be

2​K​

2K​

2K

K

Kinetic energy of the ball =K and angle of projection (θ)=45∘. Velocity of the ball at the highest point =vcosθ=vcos45∘=2​v​. Therefore kinetic energy of the ball =21​m×(2​v​)2=41​mv2=2K​.

Q. A particle is projected making an angle of $4 5^{\circ}$ with horizontal having kinetic energy $K$ . The kinetic energy at highest point will be

$\frac{K}{2}$

$\frac{K}{2}$

$2 K$

$K$

Kinetic energy of the ball $= K$
and angle of projection $(θ) = 4 5^{\circ}$ .
Velocity of the ball at the highest point
$= v cos θ = v cos 4 5^{\circ} = \frac{v}{2}$ .
Therefore kinetic energy of the ball
$= \frac{1}{2} m \times (\frac{v}{2})^{2} = \frac{1}{4} m v^{2} = \frac{K}{2}$ .