Thank you for the opportunity to help you with your question!
The key to solve a physics problem is to do this process everytime
1-translate the problem into variable form
2- choose a formula that involves all the variables
3-Plug in the numbers
Ok here it goes
1- s =15m (distance)
vi = 3 m/s (initial velocity)
vf=0 (final velocity) this one comes from the fact that at the highest point your velocity is 0
2- you need a formula that involves all the variables (s,vi,a,vf)
3- plug in the numbers
Please let me know if you need any clarification. I'm always happy to answer your questions.
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)
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)
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