Question

On the earth, an astronaut throws a ball straight upward; it stays in the air for a total time of 2.5 s before reaching the ground again. If a ball were to be thrown upward with the same initial speed on the moon, how much time would pass before it hit the ground? On the earth, an astronaut throws a ball straight upward; it stays in the air for a total time of 2.5 before reaching the ground again. If a ball were to be thrown upward with the same initial speed on the moon, how much time would pass before it hit the ground? 6.1 s 37 s 90 s 15 s

Answer 1

Answer: 15 s

Step-by-step explanation:

This is a situation related to vertical motion with constant acceleration, where the equation that will be usefull is:

`y=y_(o)+V_(o)t-(1)/(2)gt^(2)` (1)

Where:

`y=0` is the final height of the ball (when it reaches the ground)

`y_(o)=0` is the initial height of the ball (is also zero because is thrown from ground)

`V_(o)` is the initial velocity of the ball

`t=2.5 s` is the time the ball is in air (on Earth)

`g_(E)=9.8 m/s^(2)` is the acceleration due to gravity on Earth

`g_(M)=1.62 m/s^(2)` is the acceleration due to gravity on the Moon

Having this clear, let's solve (1) for the Earth:

`0=0+V_(o)t-(1)/(2)gt^(2)` (2)

`0=t(V_(o)-(1)/(2)gt^(2))` (3)

`V_(o)=(1)/(2)gt^(2))` (4)

`V_(o)=(1)/(2)(9.8 m/s^(2))(2.5 s)^(2))` (5)

`V_(o)=12.25 m/s` (6) This is the initial velocity

Using this same velocity and equation (4) for the Moon:

`12.25 m/s=(1)/(2)(1.62 m/s^(2))t^(2))` (7)

Finally finding
`t`:

`t=15.12 s \approx 15 s`