posted by Vico on .
From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.00 m/s and angle of 20.0 degrees below the horizontal. It strikes the ground 3.00 s later. (a) If the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. (c) Find the equations for the x- and y-components of the position as functions of time. (d) How far horizontally from the base of the building does the ball strike the ground? (e) Find the height from which the ball was thrown. (f) How long does it take the ball to reach a point 10.0 m below the level of launching?
They practically tell you how to do this problem by the order in which they ask the questions. First write down the starting conditions, then write down the equations of motion, and then solve them.
a) x = 0; y = y0
b) Vxo = 8.00 cos 20; Vyo = 8.00 sin 20
c) x = Vxo*t; y = y0 + Vyo*t - (g/2)t^2,
where g = 9.8 m/s^2
d) Vxo* 3.00 s
e) Solve the equation y = 0. Use the equation in part (c)
It will tell you what y0 is.
f) Solve the equation y = 10.0