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 19.0 m/s. The cliff is h = 24.0 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?
s

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

Explanation / Answer

There is no acceleration in the x-direction and Viy=0m/s so you use h=Viy(t)+.5(a)(t^2) -26=0-4.9(t^2) t=sqrt(-26/-4.9)=2.3s b)find out what the final y-velocity is Vf=Vi+at Vf=0+(-9.8)(2.3)=-22.54 so to find the angle you know that Vfx=16m/s b/c there is no acceleration in x-direction and Vfy=-22.54 so tan(theta)=Vfx/Vfy theta=arctan(16/-22.54)=-35.369 or 35.369 degrees below the horizontal

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