Question

1) A ball, initially at rest, is launched upward with a 20N force acting for 2 seconds. The ball flies vertically upward to a maximum height and falls back, sinking into the soft ground a distance of 10.0 cm.

Calculate the velocity after 2 seconds of force is applied, the maximum height if the force terminated when the ball was 1m above ground, the velocity of impact when it reached the ground and the average force exerted by the ground on the ball.

... so far i have 19.62 m/s after the 2 seconds of force is applied

Answer 1

what is the distance height relative to? the beginning point or the max height?

Answer 2

Look at the basic Kinematic equation:

\[v _{f}^{2}=v _{i}^{2}+2ax\]

you just calculated your initial velocity (19.62) and you know at the max height final velocity will be zero. Acceleration is g=-9.8 m/s^2. Solve for x, and don't forget to add the initial x (1m). You can employ the same equation to calculate the landing velocity. Is there a mass provided? Or a force v. time graph? I believe these would be essential in finding the average force.

Answer 3

how did you get velocity initial to be 19.62?

Answer 4

How did you find 19.62m/s

Answer 5

My velocity after 2 seconds was 1 m/s. (Delta)x = (1/2)vt. 1m = (1/2)v(2sec). (1/2) and the 2 cancel and you get 1m/s = v.

Answer 6

Using spemurphy's equation I got my max height as 1.0510 meters.