From the window of a building a ball is tossed from a height

From the window of a building, a ball is tossed from a height y_0 above the ground with an initial velocity of 8.10 m/s and angle of 21.0 degree below the horizontal. It strikes the ground 6.00 s later. 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? (Use the following as necessary: y_0. Assume SI units. Do not substitute numerical values; use variables only.) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. Find the equations for the x- and y-components of the position as functions of time. (Use the following as necessary: y_0 and t. Assume SI units.) How far horizontally from the base of the building does the ball strike the ground? Find the height from which the ball was thrown. How long does it take the ball to reach a point 10.0 m below the level of launching?

Solution

a) Vertical component of velocity = 8.1Sin 21= 3.091m/s
a = 9.81 m/s^2; vi = 3.091 m/s; t = 6s; s = ?; vf = ?
s = vit + 1/2 at^2 = (3.091 x 6) + (1/2 x 9.81 x 36)

s = 12.472 + 78.48
s = 195.126m = height of building.
x = 0, y = 195.126m

b) Horizontal component = 8.1Cos 21 = 7.562 m/s
vi, x = + 7.562 m/s
vi, y = - 3.091 m/s

c) velocity = distance /time
Distance = Vel x time

x = + 7.562 t m
y = - ( 3.091 + 9.81 t) m

d) At 6 sec:, x = 6 x 7.562 = 45.372 m

e) Height of drop = 195.126 m (found in a)

f) s = ut + 1/2 at^2
10 = 3.091 t + 4.905 t^2
4.905 t^2 + 3.091 t - 10 = 0

solving t =1.147 s