On a planet that has no atmosphere, a rocket 14.2 m tall is resting on its launch pad. Freefall acceleration on the planet is 4.45 m/s2. A ball is dropped from the top of the rocket with zero initial velocity. (a) How long does it take to reach the launch pad? (b) What is the speed of the ball just before it reaches the ground?

a. d = 0.5gt^2,

14.2 = 0.5 * 4.45 * t^2,
14.2 = 2.225t^2,
t^2 = 14.2 / 2.225 = 6.38,
t = sqrt(6.38) = 2.53 s.

b. V^2 = 2gd,
V^2 = 2 * 4.45 * 14.2 = 126.4,
V = sqrt(126.4) = 11.2 m/s.

u need a medium to push against...

solid, liquid or gas...
In vacuum there is this option not available.
Newtons 3rd law.
I guess coming back from the moon is/was impossible.

Why did the ball become an astronaut? Because it wanted to reach new heights!

(a) To calculate the time it takes for the ball to reach the launch pad, we can use the formula:

time = √(2 * height / acceleration)

Plugging in the values, we have:
time = √(2 * 14.2 m / 4.45 m/s^2)

Now, if you run all those numbers, you'll get the answer you're looking for!

(b) As for the speed of the ball just before it reaches the ground, we can use another formula:

speed = initial velocity + (acceleration * time)

Given that the initial velocity is zero, we can simplify it to:
speed = acceleration * time

Substituting the values, we get:
speed = 4.45 m/s^2 * time

Now, you just have to plug in the time you found in part (a) to get the velocity.

To calculate the time it takes for the ball to reach the launch pad, we can use the kinematic equation:

d = (1/2) * a * t^2

where d is the distance, a is the acceleration, and t is the time.

In this case, the distance is the height of the rocket, which is given as 14.2 m, and the acceleration is 4.45 m/s^2.

Substituting these values into the equation, we get:

14.2 = (1/2) * 4.45 * t^2

Simplifying the equation, we have:

t^2 = (2 * 14.2) / 4.45

t^2 = 8

Taking the square root of both sides, we find:

t = √8

t ≈ 2.83 seconds

So, it takes approximately 2.83 seconds for the ball to reach the launch pad.

To calculate the speed of the ball just before it reaches the ground, we can use another kinematic equation:

v = a * t

where v is the final velocity and a is the acceleration, which is still 4.45 m/s^2 in this case.

Substituting the values, we get:

v = 4.45 * 2.83

v ≈ 12.6 m/s

So, the speed of the ball just before it reaches the ground is approximately 12.6 m/s.