Posted by Cassandra on Saturday, September 12, 2009 at 1:10pm.
The horizontal-component, ux, of the initial velocity is the horizontal distance (195m) divided by 6.9s.
The vertical component, uy, can be found by the equation:
S = uy t + (1/2)(-g)t²
where
S is the vertical distance of the target (135 m)
uy = vertical component of velocity
t = 6.9 seconds
g = acceleration due to gravity (9.81 m/s/s/)
from which uy can be found.
The direction is atan2(uy,ux) from the horizontal.
I don't understand how to get the magnitude. Do you just add the two numbers together?
What do you do to the angle if it asks for the angle below the positive x-axis?
You can obtain the magnitude by doing a vectorial sum, namely magnitude=√(ux²+uy²)
The direction from the x-axis is obtained from tan-1(uy/ux), which goes from -π/2 to +%pi/2. Since the angle could range from 0 to 2π, it is necessary to observe the sign of ux and uy and determine the quadrant accordingly.
For the given question, the angle should be above the horizon (x-axis). If it is below, it will never reach the top of the cliff, and the solution is not correct.
If another questions requires a solution
below the x-axis, the angle will be negative between 0 and π/2.
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