Question

A sprinkler mounted on the ground sends out a jet of water at a 35 ∘ angle to the horizontal. The water leaves the nozzle at a speed of 10 m/s. How far does the water travel before it hits the ground?

This is my work:

d=(10)sin2(35)/9.81

but I do not know how to put the 2 in front of the angle after sin in my calculator; therefore, my answer has been incorrect.

Answer 1

Is this what you want to calculate?
10*sin(2*35)*1/9.81

Lets look:
time in air: hf=hi+vi*sin35*t-4.9t^2
t(10*sin35deg - 4.9t)=0
timeinair= 10sin35/9.8

now, horizontal distance:
distance=10cos35*timeinair
distance=10*cos35*10*sin35/9.81
= 100cos35deg*sin35deg *1/9.81
which is easy to put in a calulator. You may go to the double angle formulas if you wish, but honestly, it is not going to save much.

I get a little less than 5 meters