A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 9.40 m/s at an angle of 21.0Â° below the horizontal. It strikes the ground 6.00 s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
(b) Find the height from which the ball was thrown.
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?
If the height is h, then the vertical position y at time t is
y = h - (9.4 sin21°)t - 4.9t^2
y(6) = 0, so you can find h.
the horizontal speed is constant: 9.4 cos21°
Now you can answer the questions.posted by Steve