From the window of a building, a ball is tossed from a height y 0 above the grou

Question

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.40 m/s and angle of 16.0° below the horizontal. It strikes the ground 4.00 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? (Use the following as necessary: y0. Assume SI units. Do not substitute numerical values. Use variables only.)

(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. (Use the following variables as necessary: y0 and t. Assume SI units.)

(d) How far horizontally from the base of the building does the ball strike the ground?
__________ m

(e) Find the height from which the ball was thrown.
_______________m

(f) How long does it take the ball to reach a point 10.0 m below the level of launching?

_________________s

Explanation / Answer

a)

Vertical component of velocity = 8.4Sin(16) = 2.32 m/s

a = 9.81 m/s^2;

vi = 2.32 m/s;

t = 4s;

s = ?; vf = ?

s = vit + 1/2 at^2

s = (2.32 x 4) + (1/2 x 9.81 x 16)

s = 9.28 + 78.48

s = 87.76m = height of building.

x = 0, y = 87.76m

b)

Horizontal component = 8..4Cos(16) = 8.1 m/s

vi, x = + 8.1 m/s

vi, y = - 2.32 m/s

c)

velocity = distance /time

Distance = Vel x time

x = + 8.1 t m
y = - ( 2.32 + 9.81 t) m

d)

At 4 sec:, x = 4 x 8.1 = 32.4 m

e)

Height of drop = 87.76 m (found in a)

f)

s = ut + 1/2 at^2

10 = 2.32 t + 4.905 t^2

4.905 t^2 + 2.32 t - 10 = 0

t = 1.212 sec

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.