An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15.0 meters if her initial speed in 3.00 m/s. What is the free-fall acceleration on the planet?

ok I did not see the word horizontal initially that makes a difference

for the maximum horizontal distance you need 45degree shooting so if the velocity is 3m/s she will go 3sin45 vertical and 3cos45 horizontal

now distance (s) is velocity (v) multiply by time (t)

s=V.t

so s =3cos45.t

the time she has is the time her vertical velocity will buy her

t=(Vf-Vi)/a (final velocity -initial velocity)/ acceleration for only going up which is the same as time to come down

t=2*3sin45/a notice that we multiplied it by 2 to account for the time to go up and back to the ground (all the time she goes up and comes down she moves horizontally)

so s=2*3*3*sin45*cos45/a

or 15=18/2a

so a=0.6

Best

Shirin

Sep 24th, 2015

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