A cricket ball is hit at 30^{circ} with the h | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The initial kinetic energy is $K = \frac{1}{2}mv^2$,where $v$ is the initial velocity.At the highest point of the trajectory,the vertical componen

Vedclass · Accepted Answer

$3K/4$

Vedclass · Answer

(D) The initial kinetic energy is $K = \frac{1}{2}mv^2$,where $v$ is the initial velocity.At the highest point of the trajectory,the vertical component of velocity becomes zero,and the velocity is only the horizontal component,$v_x = v \cos 	heta$.The kinetic energy at the highest point is $K' = \frac{1}{2}m(v \cos 	heta)^2$.Substituting $K = \frac{1}{2}mv^2$,we get $K' = K \cos^2 	heta$.Given $	heta = 30^{\circ}$,we have $K' = K \cos^2(30^{\circ}) = K \left( \frac{\sqrt{3}}{2} ight)^2 = \frac{3K}{4}$.

$A$ cricket ball is hit at $30^{\circ}$ with the horizontal with kinetic energy $K$. The kinetic energy at the highest point is

Similar Questions

$A$ ball is rolled off the edge of a horizontal table at a speed of $4\, m/s$. It hits the ground after $0.4\, s$. Which statement given below is true?

For a given velocity,a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then:

$A$ shell fired from the base of a mountain just clears it. If $\alpha$ is the angle of projection,then the angular elevation of the summit $\beta$ is

If a particle of mass $m$ is projected at an angle $\alpha$ with the horizontal,then :

Vedclass Test Series

Exam Paper Generator

Online Exam Module