A cart is moving horizontally along a straigh | vedclass

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

Question

(C) The time taken by the cart to cover $80\,m$ is given by $t = \frac{d}{v} = \frac{80}{30} = \frac{8}{3}\,s$.For the projectile to return to the sam

Vedclass · Accepted Answer

$\frac{40}{3}\,m/s$

Vedclass · Answer

(C) The time taken by the cart to cover $80\,m$ is given by $t = \frac{d}{v} = \frac{80}{30} = \frac{8}{3}\,s$.For the projectile to return to the same point on the cart,it must be fired vertically relative to the cart. The total time of flight $T$ must be equal to the time taken by the cart to travel $80\,m$,so $T = \frac{8}{3}\,s$.The time taken to reach the maximum height is $t_h = \frac{T}{2} = \frac{8/3}{2} = \frac{4}{3}\,s$.Using the equation of motion $v = u + at$,where $v = 0$ at maximum height,$a = -g = -10\,m/s^2$,and $t = \frac{4}{3}\,s$:$0 = u - 10 	imes \frac{4}{3}$$u = \frac{40}{3}\,m/s$.Thus,the projectile must be fired with a velocity of $\frac{40}{3}\,m/s$ relative to the cart.

Similar Questions

$A$ particle reaches its highest point when it has covered exactly one half of its horizontal range. The corresponding point on the displacement-time graph is characterized by

The figure shows four paths for a kicked football. Ignoring the effects of air on the flight,rank the paths according to the initial horizontal velocity component,highest first.

$A$ projectile is thrown into space so as to have a maximum possible horizontal range of $400 \, m$. Taking the point of projection as the origin,the coordinates of the point where the velocity of the projectile is minimum are

Vedclass Test Series

Exam Paper Generator

Online Exam Module