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Author Question: A projectile is launched from ground level to the top of a cliff which is 195 away and 155 high.? (Read 1569 times)

aero

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on: Jun 23, 2013
If the projectile lands on top of the cliff 8.4  after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance.

Find theta. The degree at which this occurred. I am so confused.


TI

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Reply #1 on: Jun 23, 2013
Horizontal:
x = 195 m
t = 8.4 s
195 = vocos(theta)(8.4)
vocos(theta) = 23.214 m/s

Vertical:
y = vot + 0.5gt^2
155 = vosin(theta)(8.4) - 0.5(9.8)(8.4)^2
vosin(theta)(8.4) = 500.744
vosin(theta) = 59.612 m/s

To find the angle:
vosin(theta)/vosos(theta) = 59.612/23.214 = 2.567933
sin/cos = tan -------> inverse(tan)(2.567933) = 68.723 degrees
Substitute this into either equation:

vosin(68.723) = 59.612
vo = 59.612/sin(68.723) = 63.972
vocos(68.723) = 23.214
vo = 23.214/cos(68.723) = 63.972

initial velocity (vo) = 63.972 m/s
angle (theta) = 68.723 degrees


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aero

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Reply #2 on: Jun 29, 2013
Thanks :D


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