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 22.0 m/s. The cliff is h = 41.0 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
?=( )° below the horizontal

Explanation / Answer

a) x0 = 0m
y0 = 41 m

b)V0x =22 m/s
V0y =0 m/s

c)Vx = 22
Vy = -8.91t
since the direction is downward, the velocity should be negative. as velocity is displacement/ time, and displacement is negative.

d) x = 22 * t
y = 41 - 0.5*9.81*t^2
y = 41 - 4.9t^2

e) 41 = 4.9t^2
t^2 = 8.367
t = 2.89 seconds

f) Vx = 22 m/s
Vy = -9.81(2.89) = -28.351 m/s
Vf = (222 + (-28.351)2) = 35.89 m/s

since the y component is in the negative region and x component is in the positive region, the direction will be below the horizontal.
angle = tan-1(28.351/22) = 52.19 degree below the horizontal

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