A cart is moving horizontally along a straight line with constant speed 30 m / s. A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved 80 m. At what speed (relative to the cart) must the projectile be fired (. Take .g=10 m / s 2):

Question

A cart is moving horizontally along a straight line with constant speed 30 m / s. A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved 80 m. At what speed (relative to the cart) must the projectile be fired (. Take .g=10 m / s 2): (A) 10 m / s (B) 10 √8 m / s (C) (40/3) m / s (D) None of these

Tardigrade Admin · Accepted Answer

Time of motion, t =( x / u x )=(80/30)=(8/3) s Thus, 0=u sin θ t -(1/2) gt 2 ⇒ u sin θ=(40/3) ∴ u y =40/ 3 m / s

$10 m / s$

$108 m / s$

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

None of these

Time of motion,
$t = \frac{x}{u _{x}} = \frac{80}{30} = \frac{8}{3} s$
Thus, $0 = u sin θt - \frac{1}{2} g t^{2}$
$\Rightarrow u sin θ = \frac{40}{3}$
$∴ u_{y} = 40/3 m / s$

Q. A cart is moving horizontally along a straight line with constant speed 30m/s. A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved 80m. At what speed (relative to the cart) must the projectile be fired ( Take g=10m/s2):

10m/s

108​m/s

340​m/s

None of these

Time of motion, t=ux​x​=3080​=38​s Thus, 0=usinθt−21​gt2 ⇒usinθ=340​ ∴uy​=40/3m/s

$10 m / s$

$108 m / s$

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

Time of motion,
$t = \frac{x}{u _{x}} = \frac{80}{30} = \frac{8}{3} s$
Thus, $0 = u sin θt - \frac{1}{2} g t^{2}$
$\Rightarrow u sin θ = \frac{40}{3}$
$∴ u_{y} = 40/3 m / s$