From the window of a building, a ball is tossed from a height y0 above the groun

Question

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.80 m/s and angle of 23.0° below the horizontal. It strikes the ground 5.00 s later.

(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 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) initial coords of ball are (0,y0)
b) initial x velocity = v0 cos 23 = 8.8 cos 23=8.1m/s
initial y velocity = v0 sin 23 =8.8 sin 23 = 3.438m/s (in the negative direction)

c) x(t)=x0+v0(x) t +1/2 a(x) t^2

x0=0, v0(x)=8.1m/s and there is no accel in the x direction, so we have

x(t)=8.1t

y(t)=y0+v0(y)-1/2 gt^2 or y(t)=y0-3.438t-1/2 gt^2

d) we are told that the total time of travel is 5s, so we know that the x position at t=5 s is:

x(3)=8.1 x 3 =24.3 m

e) we find y0 by solving the equation:

y(3)=y0-3.438m/s(5s)-1/2(9.8m/s/s)(5s)^2...

since we know the ball hits the ground (y=0) when t=5s

so we have:

y0-17.19m-49m=66.19m

f) now, just solve for y(t)=10m, and solve for t using the y equation of motion

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