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Set 4 Problem number 21
An object is given a velocity of 10 m/s,
inclined at angle -5.001 degrees with respect to horizontal, at an altitude of 38 meters.
- How far will the object travel in the horizontal
direction as it falls to the ground, and what will be the magnitude and angle of its
velocity when it strikes the ground?
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) = 10 m/s * sin(-5.001 deg) = -.8718 m/s and
- v0x = v0 cos(`theta) = 10 m/s * cos(-5.001 deg) = 9.961 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 = -.8718 m/s, ay = -g (remember that
upward is now positive, so gravity is pulling in the negative direction), and `dsy = - 38
meters (displacement is in the negative direction).
- We easily obtain the final y velocity vyf from vyf^2 = vy0^2 + 2 a
`ds = ( -.8718 m/s )^2 + 2 * (-9.8 m/s^2) * (- 38 m) = 566 m^2/s^2.
- We see that vfy = +- -23.79 m/s. Since the final velocity will be downward
we have vfy = --23.79 m/s.
- Average y velocity will thus be ( 10 m/s + --23.79 m/s) / 2 = 16.89 m/s.
- The time required for displacement 38 meters at 16.89 m/s is `dt = `dsy / vyAve
= 38 meters / ( 16.89 m/s) = 2.249 sec.
We now see that the horizontal displacement is the product of the time `dt =
2.249 sec and the constant x velocity 9.961 m/s:
- Horizontal displacement = `dsx = 9.961 m/s * 2.249 sec = 22.4 meters.
The velocity of the projectile on striking the ground is the resultant of its x
and y velocity components vfx = 9.961 m/s and vfy = --23.79 m/s.
- The magnitude of the velocity is vf = `sqrt( ( 9.961 m/s)^2 + (-`vbls m/s)^2 ) =
25.79 m/s.
- The angle is tan^-1 (--23.79 m/s / ( 9.961 m/s) ) = 67.28 deg, or (360 +
67.28) deg = 427.2 deg.
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