Question

A paintball is shot horizontally in the positive x direction at time t after the ball is shot it is 4 cm to the right and 4 cm below its starting point at time 2 t what is the position of the ball relative to its starting point ignore air resistance

Answer 1

Answer:

At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom

Explanation:

 We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Considering the horizontal motion of paint ball

    Distance traveled during time t = 4 cm

    Initial velocity = u m/s

   Acceleration = 0 $m/s^2$

So $4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}$

Now at time 2t,

  $S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm$

  So horizontal distance traveled in time 2t = 8 cm to the right

Now considering the vertical motion of paint ball

  Distance traveled during time t = 4 cm

    Initial velocity = 0 m/s

   Acceleration = -g $m/s^2$

$4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}$

At time 2t,

     $S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm$

 So vertical distance traveled in time 2t = 16 cm to the bottom