SOLUTION: h = −16t2 + vt sin θ. we are asked to Find the height of the object after 2 seconds if v is 1,500 feet per second and theta is 30°. I think I am getting a little mix

Algebra ->  Absolute-value -> SOLUTION: h = −16t2 + vt sin θ. we are asked to Find the height of the object after 2 seconds if v is 1,500 feet per second and theta is 30°. I think I am getting a little mix      Log On


   



Question 960545: h = −16t2 + vt sin θ.
we are asked to Find the height of the object after 2 seconds if v is 1,500 feet per second and theta is 30°.
I think I am getting a little mixed up on my work. Is my set up supposed to look like this?
750=-16(2^2)+1500(2)?
or am I no where near the right track?

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
h = −16t2 + vt sin θ
h = -16(2^2) + 1500*2*(1/2)
note that sin(30) = 1/2
h = -64 + 1500
h = 1436 feet
the object reaches a height of 1436 feet after 2 seconds