The first one, directly proportional to the length that the spring is stretched.
Answer:
2.82 s
Explanation:
The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 is the starting position, 2.3 m in this case.
Vy0 is the starting speed, 13 m/s.
a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.
Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2
It will reach the ground when Y(t) = 0
0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2
-4.9 * t^2 + 13 * t + 2.3 = 0
Solving this equation electronically gives two results:
t1 = 2.82 s
t2 = -0.17 s
We disregard the negative solution. The ball spends 2.82 seconds in the air.
The car is initially traveling north at 17.7 m/s, and after 12 s, its velocity is 14.1 m/s, still due north. This means that the direction of the car has not changed, so we can already say that the direction of the acceleration is north (if the magnitude of the acceleration is positive) or south (if the magnitude of the acceleration is negative).
To find the magnitude of the average acceleration, we must calculate the ratio between the change in velocity and the time taken:

Since the acceleration is negative, it means it is in the opposite direction to the motion of the car, therefore south. Therefore, the correct answer is
b) 0.30 m/s2, south
the correct answer is no ;)