Two people are standing on a 3.12-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of air, like a hovercraft. The two people, the platform and a 3.36-kg ball are all initially at rest. One person throws the 3.36-kg ball to the other, who catches it. The ball travels nearly horizontally. Excluding the ball, the total mass of the platform and people is 119 kg. Because of the throw, this 119-kg mass recoils. How far does the platform move before coming to rest again

Displacement = 0.086m

Explanation:

Given

M = Combined mass of the two people and the platform = 119kg

m = mass of the ball = 3.36kg

v = velocity of the ball

V = Recoil velocity of the platform

t = time spent by the ball in the air

L = length of the platform = 3.12m

MV + mv = 0

The distance moved by the platform is given by; x = Vt

Where t = L/(v - V)

v - V represents the velocity of the ball relative to the platform.

Substitute L/(v - V) for t

so, x = VL/(v-V)

From (MV + mv = 0), we have

MV = -mv

V/v = -m/M

So, x = VL/(v-V) becomes

x = ((V/v)L)/(1 - V/v)

x = ((-m/M)L)/(1 + m/M)

x = -mL/(M+m)

x = -(3.36 * 3.12)/(3.36 + 119)

x = -0.08567505720823798627002

x = -0.086m

The negative sign implies that the displacement of the platform is in the opposite direction to the displacement of the ball.

The distance moved by the platform before coming to rest is 0.086m

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14