Question:

A can 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 can) must the projectile be fired? (Takes $ =10\,m/{{s}^{2}} $ )

Updated On: Jun 20, 2022
  • $10\, m/s$
  • $ 10\sqrt{8} \, m/s$
  • $ \frac{40}{3} \, m/s$
  • None of these
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The Correct Option is C

Solution and Explanation

As seen from the cart the projectile moves vertically upwards and comes back.
The time taken by cart to cover $80 \,m$
$=\frac{s}{v}=\frac{80}{30}=\frac{8}{3} s$
Here, $u=?, v=0, a=-g=10 \,m / s ^{2}$
(for a projectile going upwards)
and $t=\frac{8 / 3}{2}=\frac{4}{3} s$
From first equation of motion
$ v=u+a t $
$0=u-10 \times \frac{4}{3} $
$\therefore u=\frac{40}{3} \,m / s$
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Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration