Question

From the window of a building, a ball is tossed from a height y₀ above the ground with an initial velocity of 9.10 m/s and angle of 25.0° below the horizontal. It strikes the ground 6.00 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.)

xi=

yi=

(b)With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity.

vi,x=_________m/s

vi,x=_________m/s

(c) Find the equations for the x- and y-components of the position as functions of time. (Use the following as necessary: y₀ and t. Assume SI units.)

x=__________m

y=__________m

(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

Answer 1

a)

xi = 0

yi = yo

b)

x-component of the initial velocity

v0x = v0 cos25

y-component,

v0y = v0 sin25

c)

X component of position, dx = 8.247 t

y- component,

dy = yo - 3.846 t - (1/2)gt^2

d)

x = (v0 cos25)(t) = (9.1)( cos25)(6) = 49.48 m   

e)

dy = Vo yt + (1/2)gt^2

h = (9.1)(sin25) + (1/2)(9.8)(6)^2 = 180.25 m

f)

h = 10 m

10 = (9.1) (sin25)t + (1/2)(9.8)t^2   

4.9t^2 + 3.84 t - 10 = 0

t = 1.1 s

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