1. No category

14 Projectile Motion.notebook
March 10, 2015
Need to know: Kinematic Equations of Motion
Goals:
Talk about Quiz ­ Will return tomorrow
Lesson ­ Projectile Motion
Return Unit Test Part 1
1
14 Projectile Motion.notebook
March 10, 2015
Projectile Motion
Projectile = an object that has been launched and which does not have its own propulsion system
A projectile follows a parabolic path. ­­> Attempt Back of 1st Handout to see this
If it wasn’t for Earth's gravity, a projectile would have a constant velocity. The range is the distance the projectile covers in a horizontal direction. At the same time, gravity causes it to accelerate toward the ground at a rate of 9.80 m/s2. Observe the photo in textbook on p. 83. 2
14 Projectile Motion.notebook
March 10, 2015
The vertical velocity is not constant (non­uniform motion) so the kinematic equations must be used for analyzing projectile motion.
The horizontal velocity is constant (uniform motion). (Assuming air resistance is negligible)
The horizontal and vertical components of projectile motion must be calculated separately because these motions are independent of each other. 3
14 Projectile Motion.notebook
March 10, 2015
Sample problems:
1) A biathlete fires her gun at the target at the same moment that she sneezes. The bullet leaves the gun horizontally from a height of 25 cm. The bullet leaves the gun at a speed of 340 m/s and misses the target. What is its range?
4
14 Projectile Motion.notebook
March 10, 2015
2) A bowling ball is thrown horizontally from the roof of a very tall building with a velocity of 15.5 m/s [E]. a) Calculate the velocity of the ball when it hits the ground 35.5 m below its launch point.
b) Calculate the ball’s displacement with respect to its launch point after 1.50 s.
Recognize the two parts to consider: Horizontal Component and Vertical Component
• The horizontal velocity remains constant at 15.5 m/s [E]. 2
• The vertical velocity increases at 9.80 m/s .
Magnitude
Direction
The final velocity would be about 30.6 m/s [E60°Dn]
5
14 Projectile Motion.notebook
March 10, 2015
b) Calculate the ball’s displacement with respect to its launch point after 1.50 s.
To visualize the situation, arbitrarily consider the position is here after 1.50 s
Since the and components are independent of each other we can calculate displacements separately for each direction and then determine the final displacement using Trig. with the 2D vector diagram above.
Magnitude
Direction
The Displacement is about 25.7 m [E25°Dn] after 1.50 s.
6
14 Projectile Motion.notebook
March 10, 2015
Homework:
1 ­ Complete HO Projectile Motion
2 ­ WS Projectile Motion ­ "Knowledge Questions" 1­5
3 ­ Start WS Projectile Motion ­ "Thinking Questions" (Ignore #9 and #11)
Extra Practice: Textbook: p.130 Q. 20 – 24, 27
7
14 Projectile Motion.notebook
March 10, 2015
8