Question

On Mars, gravity pulls objects downwards at a constant acceleration of 12 ft/s2. Anastronaut on Mars throws a ball up into the air from some unknown initial height. After 2 secondshave passed it is 18 feet up in the air, and after 3 seconds, it hits the ground.From what height was the ball thrown?
What is the maximum height the ball achieves between the time when it is thrown into the air (t = 0)and halfa second later (t =1/2）？

Answer 1

To find the height from which the ball was thrown, we need to determine the initial velocity of the ball.

Given:
Acceleration due to gravity on Mars = 12 ft/s^2
Time after which the ball is 18 feet up = 2 seconds

Using the equation:
S = ut + (1/2)at^2

where:
S = Distance traveled
u = Initial velocity
t = Time
a = Acceleration due to gravity

For the first part of the problem, we can use the information given to solve for the initial velocity:

18 = u * 2 + (1/2) * 12 * (2^2)
18 = 2u + 12 * 4
18 = 2u + 48
2u = -30
u = -30/2
u = -15 ft/s

Therefore, the ball was thrown with an initial velocity of -15 ft/s.

For the second part of the problem, we need to find the maximum height the ball achieves between t = 0 and t = 1/2 seconds.

Using the equation:
v = u + at

where:
v = Final velocity
u = Initial velocity
t = Time
a = Acceleration due to gravity

The final velocity at t = 1/2 is:

v = -15 + 12 * (1/2)
v = -15 + 6
v = -9 ft/s

To find the maximum height, we need to find the distance traveled during this time. Using the equation:

S = ut + (1/2)at^2

where:
S = Distance traveled
u = Initial velocity
t = Time
a = Acceleration due to gravity

S = -15 * (1/2) + (1/2) * 12 * (1/2)^2
S = -15 * (1/2) + 6 * (1/4)
S = -15/2 + 6/4
S = -15/2 + 3/2
S = -12/2
S = -6 ft

Therefore, the maximum height the ball achieves between t = 0 and t = 1/2 seconds is 6 ft above the initial height.