Answer:
9.4 m
Explanation:
We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.
If the ballon is launched at 9.7 m/s at 39 degrees of elevation.
Vx0 = 9.7 * cos(39) = 7.5 m/s
Vy0 = 9.7 * sin(39) = 6.1 m/s
If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 = 0
a = -9.81 m/s^2
It will fall when Y(t) = 0
0 = 6.1 * t - 4.9 * t^2
0 = t * (6.1 - 4.9 * t)
t1 = 0 (this is when the balloon was launched)
0 = 6.1 - 4.9 * t2
4.9 * t2 = 6.1
t2 = 6.1 / 4.9 = 1.25 s
The distance from the car will be the horizonta distance it travelled in that time
X(t) = X0 + Vx0 * t
X(1.25) = 7.5 * 1.25 = 9.4 m
"Traditionally, physical science courses are categorized into four areas: physics, chemistry, earth sciences, and space sciences. It is important to understand that the arbitrary divisions of specialized knowledge are integrated into a basic overview of the physical laws that govern our universe." - Cameron University
A jog because it helps to get the muscles moving and your heart pumping blood around your body.
I hope this helps.
A) Net horizontal force:
Fx= -35 + 25 + -10N
Net vertical force
Fy= 50 - 35 = 15N
Net force
F^2 = fX^2 +Fy^2
F= 18N
B.) Acceleration of system
a=18/5=3.6m/s^2
C). Equilibriant force must be equal to the net force above
F1= 18N
D.) Acceleration becomes half
a1= a/2 =1.8m/s^2
E.) Equilibrant force also doubles
F" = 2F' = 36N
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