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 p…

Question

asked Mar 3, 2019 61.3k views

1 Answer

Petr Broz · Answer 1 · 2019-03-07T03:23:08+0000

Answer:

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

Step-by-step explanation:

We have equation of motion ,
s= ut+(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
$4 = u*t+(1)/(2) *0*t^2\\ \\ u = (4)/(t)$

Now at time 2t,

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

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

$4=0*t-(1)/(2) *g*t^2\\ \\ t^2=(-8)/(g)$

At time 2t,

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

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