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 the ball to reach the launch pad?
b. What is the speed of the ball just before it hits the
ground? The velocity?
c. What is the 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.
c. What is the velocity of the ball half way down?
To solve the given problem, we can use the equations of motion. Let's consider the motion of the ball when it is dropped from the top of the rocket.
a. To find the time it takes for the ball to reach the launch pad, we can use the equation:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity, a is acceleration, and t is time.
In this case, the initial velocity of the ball is zero (u = 0), the distance traveled is 14.2 m (s = 14.2 m), and the acceleration is 4.45 m/s^2 (a = 4.45 m/s^2).
We can rearrange and solve for time:
14.2 = 0.5 * 4.45 * t^2
Simplifying the equation:
14.2 = 2.225 * t^2
Divide both sides by 2.225:
t^2 = 14.2 / 2.225
t^2 = 6.38
Taking the square root of both sides:
t = √(6.38)
t ≈ 2.527 seconds
Therefore, it takes approximately 2.527 seconds for the ball to reach the launch pad.
b. To find the speed of the ball just before it hits the ground, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is acceleration, and t is time.
In this case, the initial velocity of the ball is zero (u = 0), the acceleration is 4.45 m/s^2 (a = 4.45 m/s^2), and we already know the time it takes for the ball to reach the launch pad is approximately 2.527 seconds.
Substituting the given values:
v = 0 + 4.45 * 2.527
v ≈ 11.26 m/s
Therefore, the speed of the ball just before it hits the ground is approximately 11.26 m/s.
c. To find the velocity of the ball halfway down, we can use the equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is acceleration, and s is the distance traveled.
In this case, the initial velocity of the ball is zero (u = 0), the acceleration is 4.45 m/s^2 (a = 4.45 m/s^2), and the distance traveled halfway down is half the total distance, which is 14.2 m / 2 = 7.1 m (s = 7.1 m).
Substituting the given values:
v^2 = 0 + 2 * 4.45 * 7.1
v^2 ≈ 62.995
Taking the square root of both sides:
v ≈ √(62.995)
v ≈ 7.95 m/s
Therefore, the velocity of the ball halfway down is approximately 7.95 m/s.