A student stands at the edge of a cliff and throws a stone horizontally over the

Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 17.0 m/s. The cliff is h = 29.0 m above a flat, horizontal beach as shown in the figure.

(a) What are the coordinates of the initial position of the stone?
x0 = 1 m
y0 = 2 m


(b) What are the components of the initial velocity?
v0x = 3 m/s
v0y = 4 m/s

(c) Write the equations for the x- and y-components of the velocity of the stone with time. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.)
vx =
vy =

(d) Write the equations for the position of the stone with time, using the coordinates in the figure. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.)
x =
y =


(e) How long after being released does the stone strike the beach below the cliff?
9 s

(f) With what speed and angle of impact does the stone land? vf = 10 m/s
? = 11° below the horizontal

Explanation / Answer

a) x0=0m

y0=29m

b)v0x=17 m/s

v0y=0m/s

c)vx=v0=17

vy= -at = -9.8t

d)x=x0+v0*t+0.5*a*t2 = 17*t

y=y0+v0*t + 0.5*a*t2 = 29 -0.5*gt2= 29-4.9t2

e) -29 = -4.9t2 ----> t = 2.43s

f) vx = 17 m/s

vy = -9.8*2.43=23.81 m/s

v = (vx2+vy2)=29.26 m/s

=tan-1(23.81/17)=54.47 degrees

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