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

Explanation:

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

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (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:  

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (2)  

$0=t(V_{o}-\frac{1}{2}gt^{2})$ (3)  

$V_{o}=\frac{1}{2}gt^{2})$ (4)  

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2.5 s)^{2})$ (5)  

$V_{o}=12.25 m/s$ (6)  This is the initial velocity

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

$12.25 m/s=\frac{1}{2}(1.62 m/s^{2})t^{2})$ (7)  

Finally finding $t$:

$t=15.12 s \approx 15 s$