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Set 4 Problem number 21
How far will an object travel in the horizontal
direction as it falls freely to the ground from an altitude of 40 meters, and what will be
the magnitude and angle of its velocity when it strikes the ground, if at this altitude it
is given a velocity of 9 m/s, inclined at angle -37.01 degrees with respect to
horizontal.
We consider the horizontal and vertical motions, both of which are subjected to
uniform acceleration, separately.
Using the upward (y) direction and the rightward (x) direction as positive, we
have initial vertical and horizontal velocities
- v0y = v0 sin(`theta) = 9 m/s * sin(-37.01 deg) = -5.418 m/s and
- v0x = v0 cos(`theta) = 9 m/s * cos(-37.01 deg) = 7.186 m/s.
We first analyze the vertical motion to find the final vertical velocity and the
time required to reach the ground:
- The vertical motion is characterized by v0y = -5.418 m/s, ay = -g (remember that
upward is now positive, so gravity is pulling in the negative direction), and `dsy = - 40
meters (displacement is in the negative direction).
- We easily obtain the final y velocity vyf from vyf^2 = vy0^2 + 2 a
`ds = ( -5.418 m/s )^2 + 2 * (-9.8 m/s^2) * (- 40 m) = 23010 m^2/s^2.
- We see that vfy = +- -151.6 m/s. Since the final velocity will be downward
we have vfy = --151.6 m/s.
- Average y velocity will thus be ( 9 m/s + --151.6 m/s) / 2 = 80.3 m/s.
- The time required for displacement 40 meters at 80.3 m/s is `dt = `dsy / vyAve
= 40 meters / ( 80.3 m/s) = .4981 sec.
We now see that the horizontal displacement is the product of the time `dt =
.4981 sec and the constant x velocity 7.186 m/s:
- Horizontal displacement = `dsx = 7.186 m/s * .4981 sec = 3.579 meters.
The velocity of the projectile on striking the ground is the resultant of its x
and y velocity components vfx = 7.186 m/s and vfy = --151.6 m/s.
- The magnitude of the velocity is vf = `sqrt( ( 7.186 m/s)^2 + (-`vbls m/s)^2 ) =
151.7 m/s.
- The angle is tan^-1 (--151.6 m/s / ( 7.186 m/s) ) = 87.28 deg, or (360 +
87.28) deg = 447.2 deg.
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