While exploring Mars, an astronaut launches a ball straight up from the Martian
ID: 1577376 • Letter: W
Question
While exploring Mars, an astronaut launches a ball straight up from the Martian surface with a vertical velocity of 10m/s and notices that it returns to the starting vertical position after 5.198 seconds. He has another ball-launcher that he suspects is mislabeled and does not launch projectiles at the stated 25 m/s. To test this idea, he launches a ball at 600 above the Martian surface and notices that it lands 168 meters (horizontally) from the launcher. Is the label on the ball launcher incorrect and if so, by how much?
Explanation / Answer
let g is the acceleration due to gravity on Mars.
we know time of flight, t = 2*vo/g
==> g = 2*vo/t
= 2*10/5.198
= 3.8476 m/s^2
when the ball is projected horizontally above 600 m from the surfcae.
let t is the time taken to land.
use, y = (1/2)*g*t^2
t = sqrt(2*y/g)
= sqrt(2*600/3.8476)
= 17.66 s
now use,
speed of the ball, vo = x/t
= 168/17.66
= 9.51 m/s
so, the lable on the launcher is incorrect.
difference = 25 - 9.51
= 15.5 m/s
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