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 , [tex]s= ut+\frac{1}{2} at^2[/tex], 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 [tex]m/s^2[/tex]
So [tex]4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}[/tex]
Now at time 2t,
[tex]S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm[/tex]
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 [tex]m/s^2[/tex]
[tex]4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}[/tex]
At time 2t,
[tex]S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm[/tex]
So vertical distance traveled in time 2t = 16 cm to the bottom
