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 = 72.0 m above a flat, horizontal beach as shown in the figure.

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

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


(f) With what speed and angle of impact does the stone land?
vf = m/s
? = ° 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 = 72 -0.5*gt2= 72-4.9t2
e) -72 = -4.9t2 ----> t = 3.83s
f) vx = 17 m/s
vy = -9.8*3.83=37.53 m/s
v = (vx2+vy2)=41.20 m/s
=tan-1(41.20/17)=67.58 degrees

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