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 8.90 m/s and angle of 25.0° below the horizontal. It strikes the ground 5.00 s later.

(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. Let the variable t be measured in seconds.)
x = Your answer is incorrect. 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

Explanation / Answer

we'll use the equations:

x=x0+v0xt+0.5axt2

y=y0+v0yt+0.5ayt2

ax=0

ay=g=-9.81[m/s2]

x0=0

v0=8.9[m/sec]

=-25o ---> v0x=v0cos=8.066[m/s]

v0y=v0sin=-3.761[m/s]

t=5

c.

x=8.066t

y=y0-3.761t-0.5*9.81t2=y0-3.761t-4.905t2

d.

since t=5[s] then: x=8.066*5=40.33[m]

e.

when the ball hits the ground, it's height is 0, so we get:

0=y0-3.761*5-4.905*52 --> y0=3.761*5+4.905*52=141.43[m]

f.

since y0=141.43[m], then 10 meteres below that is: 131.43[m]:

131.43=141.43-3.761t-4.905t2 ---> 4.905t2+3.761t-10=0

t1.095[s]

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