April 20th, 2011

1.)We have seen objects launched horizontally, following the trajectory shown below.

2) Now, we will look at projectiles that have been launched at an angle to the horizontal. They will rise upwards to a peak while moving horizontally.

Once the projectile has reached its peak, it will fall with a motion that is equal but opposite to its path upwards to the peak.

A non-horizontally launched projectile is a projectile that begins its motion with an initial velocity that is both horizontal and vertical.

We must use our trig functions and SOHCAHTOA to determine the horizontal and vertical components of the projectile.

For an object launched as shown.

A football is kicked with an initial velocity of 25 m/s at an angle of 40-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.

Horizontal Component vix = vi*cos vix = 25 m/s*cos 40vix = 19.2 m/sa = 0m/s

Vertical Component viy = vi*sin viy = 25 m/s*sin 40 Viy = 16.1 m/sa = -9.81m/s2

vfy = viy + ayt (use this equation and solve for time)-16.1 m/s = 16.1 m/s + (-9.81 m/s2)t-32.2 m/s = (-9.81 m/ s2)t 3.28 s = t

The total time of flight of the football is 3.28 seconds.

dx = vixt (use this equation to look for the horizontal displacement of the projectile)

dx = (19.2 m/s) (3.28 s)

dx = 63.0 m

To find the vertical displacement (when t=1.64s)

dy = viyt +0.5ayt2

dy = (16.1 m/s) (1.64 s) + 0.5 (-9.81 m/ s2) (1.64 s)2

dy = 26.4 m + (-13.2 m) dy = 13.2 m

1. Use the given values of the initial velocity (the magnitude and the angle) to determine the horizontal and vertical components of the velocity

2. List known and unknown information in terms of the symbols of the kinematic equations. (keep horizontal and vertical givens separate)

3. Identify the unknown quantity that the problem requests you to solve for.

4. Select either a horizontal or vertical equation to solve for the time of flight of the projectile.

5. With the time determined, use a horizontal equation to determine the horizontal displacement of the projectile.

6. Finally, the peak height of the projectile can be found using a time value that is one-half the total time of flight.

Pg. 543 #’s 9, 10, 11, 12