A ball is projected from the ground at an ang | vedclass

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

Question

(A) The velocity components of a projectile at time $t$ are $v_x = u \cos 	heta$ and $v_y = u \sin 	heta - gt$.The kinetic energy $K$ is given by $K

Vedclass · Accepted Answer

Vedclass · Answer

(A) The velocity components of a projectile at time $t$ are $v_x = u \cos 	heta$ and $v_y = u \sin 	heta - gt$.The kinetic energy $K$ is given by $K = \frac{1}{2} m v^2 = \frac{1}{2} m (v_x^2 + v_y^2)$.Substituting the components,$K = \frac{1}{2} m [(u \cos 	heta)^2 + (u \sin 	heta - gt)^2]$.Expanding this,$K = \frac{1}{2} m [u^2 \cos^2 	heta + u^2 \sin^2 	heta + g^2 t^2 - 2 u g t \sin 	heta] = \frac{1}{2} m [u^2 + g^2 t^2 - 2 u g t \sin 	heta]$.This is a quadratic equation in $t$ of the form $K = at^2 + bt + c$,which represents a parabola opening upwards.At $t = 0$,$K = \frac{1}{2} m u^2$. At the maximum height $(t = \frac{u \sin 	heta}{g})$,the vertical velocity is zero,so $K$ is minimum but non-zero $(K_{min} = \frac{1}{2} m (u \cos 	heta)^2)$.Thus,the graph is a parabola that decreases until the maximum height and then increases,which matches the shape in Graph $A$.

$A$ ball is projected from the ground at an angle of $\theta$ from the horizontal. The graph of kinetic energy $(K)$ versus time $(t)$ will be:

Similar Questions

$A$ body is projected from the ground at an angle $\theta_1$ with the horizontal with kinetic energy $K$. What is the potential energy of the body when the velocity makes an angle $\theta_2$ with the horizontal?

$A$ mass of $2 \, kg$ is whirled in a horizontal circle by means of a string at an initial speed of $5$ revolutions per minute. Keeping the radius constant,the tension in the string is doubled. The new speed is nearly ....... $rpm$.

$A$ particle is moving in the $xy$-plane in a circular path with its center at the origin. If at an instant the position of the particle is given by $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$,then the velocity of the particle is along .......

The position vector of a particle is given as $\vec{r} = (t^2 - 4t + 6)\hat{i} + (t^2)\hat{j}$. The time after which the velocity vector and acceleration vector become perpendicular to each other is equal to ....... $sec$.

$A$ particle is projected from the horizontal making an angle of $53^{\circ}$ with an initial velocity of $100\,m/s$. The time taken by the particle to make an angle of $45^{\circ}$ with the horizontal is $.........\,s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module