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Since the ball rolls of the cliff its initial vertical velocity will
be horizontal. The vertical equation of motion will then be
1
(1) y = 10m - - g t^2
2
Where g = 9.8 m/s^2 is the magnitude of the acceleration due to gravity,
t is time and we take zero altitude to be at the base of the cliff. The
ball hits when y = 0 so
1
(2) 0 = 10m - - 9.8 m/s^2 ti^2
2
Where ti is the fall time. Solving for ti
(3) ti = sqrt(2 x 10m / 9.8 m/s^2)
= 1.429 s
It takes 1.43 s to hit the ground. During this time it travels
horizontally 5 m. The horizontal velocity (and also the initial
velocity since this component does not change) is thus
(4) horizontal velocity = 5.0 m / 1.429s
= 3.499 m/s
The vertical component of velocity ( since it has no initial
vertical velocity) is given by
(5) v vertical = - gt
= -9.8m/s^2 (1.429s)
= -14.004 m/s
The magnitude of the final velocity is then
v final = sqrt( (v horizontal)^2 + (v vertical)^2)
= sqrt( (3.499 m/s)^2 + (14.004 m/s)^2)
= sqrt( 208.355 (m/s)^2)
= 14.435 m/s
the direction that the ball impacts the ground is
given by
v horizontal
------>
\ |
\ |
|v horizontal| \ | v vertical
(6) tan(a) = -------------- \ |
|v vertical| \a|
\|
v
Where a is the angle FROM vertical.
Thus
(7) tan(a) = 3.499 / 14.004
= 0.2499
thus
(8) a = 14.03 degrees
To two significant figures then
*************************************************
* (a) time of flight ~ 1.4 s *
* *
* (b) initial speed ~ 3.5 m/s (horizontal) *
* *
* (c) final velocty ~ 14 m/s at an angle *
* of 14 degrees from the *
* vertical. *
* *
*************************************************
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