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.
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
To properly input the angle squared into your calculator, you can use the following steps:
1. Convert the angle from degrees to radians: Multiply the angle (35 degrees) by π/180 to get the angle in radians. This will give you the value of the angle in radians, which is approximately 0.6109 radians.
2. Square the angle: Multiply the angle in radians by itself. In this case, square approximately 0.6109 to get approximately 0.3734.
3. Now, you can substitute the squared angle value (0.3734) into your original equation:
d = (10 * sin(2 * 0.6109)) / 9.81
This formula will give you the correct answer for the distance the water travels before it hits the ground.