Asked by Hannah

Solve in Calculus Based (Projectile)
A man stands on the roof of a 15.0 meters tall building and throws a rock with a velocity of magnitude 30 m/s at an angle of 33° above the horizontal. You can ignore air resistance. Calculate (a.) the maximum height above the roof reached by the rock; (b) the magnitude of the velocity of the rock just before it strikes the ground, and (c) the horizontal range from the base of the building to the point where the rocks strikes the ground.

Pleaseee, I don't understand. I'm gonna fail if I don't get the answers.
Answered by oobleck
using the usual equation of motion, we have the height
y = 15.0 + 30sin33° t - 4.9t^2
(a) the vertex is, as always, at t = -b/2a -- no calculus needed here
(b) v = 30sin33° - 9.8t
so, solve y=0 to get the vertical component v<sub><sub>y</sub></sub> of the velocity on impact
The horizontal part is constant v<sub><sub>x</sub></sub>=33cos33°
Then |v| = √(v<sub><sub>x</sub></sub>^2+v<sub><sub>y</sub></sub>^2)
(c) you have the speed and the time ...
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