Question

I need help with this and if you are able to find the answer pls show the steps thank you ​

Answer 1

c. 34.5 m

Explanation:

Over horizontal ground, the range R for velocity v and launch angle θ is:

R=v^2sin2θ/g

=35^2sin16∘/9.8

= 34.5 meters (so the answer is (c))