<span>a) 1960 m
b) 960 m
Assumptions.
1. Ignore air resistance.
2. Gravity is 9.80 m/s^2
For the situation where the balloon was stationary, the equation for the distance the bottle fell is
d = 1/2 AT^2
d = 1/2 9.80 m/s^2 (20s)^2
d = 4.9 m/s^2 * 400 s^2
d = 4.9 * 400 m
d = 1960 m
For situation b, the equation is quite similar except we need to account for the initial velocity of the bottle. We can either assume that the acceleration for gravity is negative, or that the initial velocity is negative. We just need to make certain that the two effects (falling due to acceleration from gravity) and (climbing due to initial acceleration) counteract each other. So the formula becomes
d = 1/2 9.80 m/s^2 (20s)^2 - 50 m/s * T
d = 1/2 9.80 m/s^2 (20s)^2 - 50m/s *20s
d = 4.9 m/s^2 * 400 s^2 - 1000 m
d = 4.9 * 400 m - 1000 m
d = 1960 m - 1000 m
d = 960 m</span>
The normal force is always perpendicular to the surface. So it would be straight to the left of the wall
Answer:
a) 4.2m/s
b) 5.0m/s
Explanation:
This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.
The problem is also an illustration of elastic collision where there is no loss in kinetic energy.
Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses 
and 
whose respective velocities before collision are 
and 
;
where 
and 
are their respective velocities after collision.
Given;
Note that =0 because the second mass was at rest before the collision.
Also, since the two masses are equal, we can say that 
so that equation (1) is reduced as follows;
m cancels out of both sides of equation (2), and we obtain the following;
a) When 
, we obtain the following by equation(3)
b) As stops moving 
, therefore,
