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 Vo=12.5m/s. The cliff is h=26.0m above a flat, horizontal beach as shown in the figure.

A) What are the coordinates of the inital position of the stone? Xo and Yo?

B) What are the components of the initail velocity? Vox and Voy?

C) Write the equations for the x and y components of the velocity of the stone with time? Vx adn Vy?

D) Write the equation for the position of the stone with time, using the coordinates in the figure? X and 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 and ?

Explanation / Answer

Here ,

v0 = 12.5 m/s

h = 26 m

A) for the initial position ,

x0 = 0 m

y0 = 26 m

B)

for the initial velocity ,

v0x = 12.5 m/s

v0y = 0 m/s

C)

for the velocity at any time

vx = v0x = 12.5 m/s

vy = 0 - gt = -9.8t

D)

X = 12.5 * t

Y = uy *t - 0.50 * g * t^2 + y0

Y = 26 -4.9 * t^2

E)
for the final Y to be zero

Y = 0 = 26 -4.9 * t^2

solving for t

t = 2.304 s

the time taken is 2.304 s

f) for the final velocity

v^2 - 12.5^2 = 2 * 9.8 * 26

solving for v

v = 25.8 m/s

the final velocity is 25.8 m/s

angle of final velocity = arctan(2.304 * 9.8/12.5) degree below horizontal

angle of final velocity = 61 degree below horizontal

angle of final velocity is 61 degree below horizontal

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