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-17.0 m/s. The ciff is h -21.0 m above a flat, horizontal beach as shown in the figure (a) What are the coordinates of the initial position of the stone? Xom Yo- (b) What are the components of the initial velocity? mý/s my's Voy (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) (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.) e) How long after being released does the stone strike the beach below the cif? (1) With what speed and angle of impact does the stone land? m/s e below the horizontal

Explanation / Answer

(a)
Considering the cliff is at y-axis, As the height of the cliff is h = 21.0 m

the coordinates of the initial position of the stone are

x0 = 0 m

y0 = 21.0 m

(b)

Given that the stone is thrown horizontally over the edge with a speed of 17.0 m/s

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

v0x = 17.0 m/s

v0y = 0 m/s

(c)

The equations of x ,y components of velocity are

vx = v0x + ax*t and vy = v0y + ay*t

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 vx = 17.0 m/s and vy = -gt = -9.8t (negative sign indicates that the velocity is along negative y-direction)

(d)

Now the equations of x and y are

x = x0 + v0x*t + (1/2)ax*t^2 and

y = y0 + v0y*t + (1/2)ay*t^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 = 17t and y = 21 - (1/2)g*t^2 = 21 - 4.9t

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

So we should have 0 = 21 - (1/2)gt^2

(1/2)(9.8 m/s^2)*t^2 = 21 m

t^2 = 4.285 s^2

t = 2.07 s

(f)

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

vx = 17.0 m/s

and vy = -(9.8 m/s^2)(2.07 s)

= -20.286 m/s

Then the magnitude of teh velocity of teh stone is

v = sqrt[(17.0 m/s)^2 + (-20.28 m/s)^2]

= 26.463 m/s

The angle of impact will be

tan^-1 (-20.28 / 17) = 50.02 degree with the negative x-axis

or 129.97 degree with the positive x-axis

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