A boy throws a ball with a velocity V_0 at an | vedclass

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

Question

(A) The horizontal component of the velocity of the ball is $V_x = V_0 \cos \alpha$.Since there is no acceleration in the horizontal direction,the bal

Vedclass · Accepted Answer

$V_0 \cos \alpha$

Vedclass · Answer

(A) The horizontal component of the velocity of the ball is $V_x = V_0 \cos \alpha$.Since there is no acceleration in the horizontal direction,the ball travels with this constant horizontal velocity throughout its flight.The time of flight $T$ is given by $T = \frac{2 V_0 \sin \alpha}{g}$.The horizontal range $R$ covered by the ball is $R = V_x 	imes T = (V_0 \cos \alpha) 	imes \left( \frac{2 V_0 \sin \alpha}{g} ight)$.To catch the ball,the boy must cover the same horizontal distance $R$ in the same time $T$.Let the velocity of the boy be $v_b$. Then,$R = v_b 	imes T$.Equating the two expressions for $R$,we get $v_b 	imes T = (V_0 \cos \alpha) 	imes T$.Therefore,the velocity of the boy must be $v_b = V_0 \cos \alpha$.

$A$ boy throws a ball with a velocity $V_0$ at an angle $\alpha$ to the ground. At the same time,he starts running with a uniform velocity to catch the ball before it hits the ground. To achieve this,he should run with a velocity of

Similar Questions

$A$ particle crossing the origin at time $t=0$,moves in the $xy$-plane with a constant acceleration $a$ in the $y$-direction. If the equation of motion of the particle is $y = bx^2$ (where $b$ is a constant),then its velocity component in the $x$-direction is

$A$ bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \; km$ away. By adjusting its angle of projection,can one hope to hit a target $5.0 \; km$ away? Assume the muzzle speed to be fixed,and neglect air resistance.

Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are $v_1$ and $v_2$ at angles $\theta_1$ and $\theta_2$ respectively from the horizontal,then the trajectory of particle $1$ with respect to particle $2$ will be:

$A$ particle of mass $m$ performs uniform circular motion of radius $r$ with linear speed $v$ under the application of force $F$. If $m$,$v$,and $r$ are all increased by $20 \%$,the necessary change in force required to maintain the particle in uniform circular motion is: (in $\%$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module