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 = 12.0 m/s. The cliff is h = 27.0 m above a flat, horizontal beach as shown in the figure 0 (a) What are the coordinates of the initial position of the stone? (b) What are the components of the initial velocity? vox = m/s 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 include units in your answer.)

Explanation / Answer

a) As the height of the cliff is h = 27.0 m, therefore the coordinates of the initial position of the stone are

x0 = 0 m and y0 = 2b7 m.

b) Given that the stone is thrown horizontally over the edge with a speed of 12.0 m/s

Therefore initially the stone has horizontal component only. and the vertical component of initial velocity is zero.

v0x = 12.0 m/s and v0y = 0 m/s

c) The equations of x y components of velocity are as shown below.

vx = v0x + axt and vy = v0y + ayt

But the horizontal acceleration of the stone is zero. Therefore ax= 0

vx = v0x + 0*t ==> vx = v0x

And the vertical acceleration of the stone is equal to -g (the free fall accelaration). Therefore ay= -g

Therefore vx = 12 m/s and vy = -gt (negative sign indicates that the velocity is along negative y-direction)

(d) Now the equations of x and y are as shown below.

x = x0 + v0xt + (1/2)axt^2 and y = y0 + v0yt + (1/2)ayt^2

But the horizontal acceleration of the stone is zero. Therefore ax= 0

And the vertical acceleration of the stone is equal to -g (the free fall accelaration). Therefore ay= -g

Therefore x = 12t and y = 27 - (1/2)gt^2

(e) When the ball strikes the ground its vertical displace must be zero. therefore y = 0

So we should have 0 = 27 - (1/2)gt^2

or (1/2)(9.8 m/s^2)t^2 = 27 m

or t^2 = 5.51 s^2

or t = 2.35 s

(f)

Now the the components of velocity of teh stone before the stone hits the ground are

vx = 12.0 m/s

and vy = -(9.8 m/s^2)(2.35 s)

= -23 m/s

Then the magnitude of teh velocity of teh stone is

v = sqrt[(12.0 m/s)^2 + (-23 m/s)^2]

= 25.95 m/s

Theangle of impact will be

tan^-1 (-23 / 12) = 62.5 with the negative x-axis or 117.5 with the positive x-axis

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