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April 24, 2014

April 24, 2014

Posted by **Cassandra** on Saturday, September 12, 2009 at 1:10pm.

- math -
**MathMate**, Saturday, September 12, 2009 at 5:52pmThe 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.

- math -
**Cassandra**, Saturday, September 12, 2009 at 7:17pmI 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?

- math -
**MathMate**, Saturday, September 12, 2009 at 11:27pmYou 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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