Warm-Up 09/13/10 Please express the Graphic Vector Addition Sums in MAGNITUDE-ANGLE format (last two pages of PhyzJob packet)

Projectile Motion Projectile Motion: motion in two dimensions (horizontal and vertical) with the vertical motion under the action of gravity only (downward). The initial velocity is in the horizontal direction. Because the action of gravity is in the vertical direction, the horizontal motion has zero acceleration if air resistance is ignored. The vertical motion is a free fall, so the acceleration due to gravity, g, is -9.80 m/s2

Because the only acceleration is that of gravity, the time it takes for the projectile to reach the ground is the same as the time it would take if the object were simply dropped.

Equations for Projectile Motion: Resolve the motion into x and y components

Example 1: A package is dropped from an airplane traveling with a constant horizontal speed of 120 m/s at an altitude of 500 m. What horizontal distance, or range, does the package travel before hitting the ground?

Solution for Example 1

Problem-Solving Tip The quantities such as initial velocities and displacements have to be treated independently. The initial horizontal velocity is 120 m/s and the initial vertical velocity is 0 m/s. 120 m/s can only be used in the horizontal motion and the 0 m/s can be used only in the vertical motion. A common mistake is to use the 120 m/s for the vertical motion.

Example 2: A golfer hits a golf ball with a velocity of 35 m/s at an angle of 25 o above the horizontal. If the point where the ball is hit and the point where the ball lands are at the same level, A) How much time does the ball spend in the air? B) What horizontal distance does the ball travel before landing?

Setting Up Example 2 Given: horizontal motion vertical motion X0=0 m y0 =0 m Vx0 = V0cosq Vy0=V0sinq y =0 m Vx0 = (35 m/s) cos 25o Vy0= (35m/s)sin25o = 31.7 m/s = 14.8 m/s Find: A) t and B) x y=y0+ Vy0t – ½ gt2, solve quadratic for t x = x0 + Vx0t

Solution A) The t=0 corresponds to the starting position and the t = 3 Solution A) The t=0 corresponds to the starting position and the t = 3.02 s corresponds to the landing position, so the time of flight is 3.0 s ( 2 sig figs) B)

Regardless of it’s path, a projectile will always follow these rules Projectiles always maintain a constant horizontal velocity (neglecting air resistance) Projectiles always experience a constant vertical acceleration of 9.8 m/s2 downward (neglecting air resistance) Horizontal and vertical motion are completely independent of one another. Therefore, the velocity of a projectile can be separated into horizontal and vertical components.

Cont. For a projectile beginning and ending at the same height, the time it takes to rise to its highest point equals the time it takes to fall from the peak back to the original height. Objects dropped from a moving vehicle have the same velocity as the moving vehicle In order to solve projectile exercises, you MUST consider horizontal and vertical motion separately.

Concept Test – slides 14-41 D:\Chapter_03\Assess\Assess_Present\WBL6_ConcepTests_Ch03.ppt Physlet Exploration 3.5 Homework: p 98 – 100; 61, 66, 67, 70, 75, 79, 81, 83, 84, BONUS: #71

Warm-Up 09/15/09 A small plane takes off at a constant velocity of 150 km/h at an angle of 37°. At 3.00 s, (a) how high is the plane above the ground, and (b) what horizontal distance has the plane traveled from the liftoff point?  

Warm-Up 09/21/09 A ball is thrown horizontally out the window of a building with a velocity of 6.5 m/s from a height of 100m. How far from the base of the building will the ball land? **Tip: find time, t, using the equation y=y0 + Vy0t – ½(gt2), then use t to calculate x by x = x0 + Vx0t