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

Question

A student stands at the edge of a cliff and throws a stone horizontally over teh edge with a speed of 19.5 m/s. The cliff is 74 meters above a flat, horizontal beach as shown in the figure

A. What are the coordinates of the initial position of the stone

x initial = m.

y iniital = m.

B. What are the cmponents of the initial velocity

Vox= m/s

Voy = m/s

C. Write the equations for the x and y components of the velocity of the stone with time. Use the following as necesary: t let the variable t be measured in seconds. Do not include 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 second, Do not state the 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 teh stone land?

Vf= m/s

angle= degrees below the horizontal.

Explanation / Answer

A)

x = 0 m

y= 74 m

B)

Vox = 19.5 m/s

Voy = 0 m/s

C)

Vx = 19.5 m/s (constant)

Vy = gt = 9.8t m/s

D)

x = 19.5t m

Y = 74 - 0.5gt2 = 74 - 4.9t2 m

E)

Y = 0

74 - 4.9t2 = 0

t = 3.09 sec

F)

after t =3.09 s

Vx = 19.5 m/s

Vy = 9.8*3.09 = 30.3 m/s

V = sqrt(Vx2 + Vy2) = 36 m/s

angle of impact = tan-1(30.3/19.5) = 57.23o below horizantal

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