A projectile is launched from ground level to the top of a cliff which is 195m away and 135m high. If the projectile lands on top of the cliff 6.9s after it is fired, find the initial velocity of the projectile ((a)magnitude and (b)direction. Neglect air resistance.

Answer 1

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.

Answer 2

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?

Answer 3

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.