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Projectiles Fired at an Angle

Projectiles Fired at an Angle

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Projectiles Fired at an Angle

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  1. v0 vx = v0cos vy = v0sin  Projectiles Fired at an Angle Now let’s find range of a projectile fired with speed v0 at an angle . Step 1: Split the initial velocity vector into components. And v0 = vx + vy

  2. v0 v0cos v0sin θ Projectiles Fired at an Angle (cont.) Step 2: Find hang time. Use y = v0t + ½at2with only vertical data:  y = (v0sin)t + ½(-a)t2 ALL THINGS VERTICAL Over level ground, y = 0. Divide through by t: 0 = v0sinθ- ½(9.8m/s2)t, Rearrange equation: t = (v0sinθ)/4.9m/s2 Note: If we had shot the projectile from a 100 m cliff,  y would be -100 m.

  3. Projectiles Fired at an Angle (cont.) ALL THINGS HORIZONTAL Step 3: Now that we know how long it’s in the air, we know how long it travels horizontally. (The projectile’s vertical and horizontal movements are completely independent.) Use Δx = v0t + ½at2again, this time with only horizontal data: Δx = (v0cos)t + ½(0)t2 = (v0cos)t v0 This is the same as saying:horiz. distance = horiz. speed × time In other words, d = vt v0sinθ θ v0cos

  4. Symmetry and Velocity - The projectile’s speed is the same at points directly across the parabola (at the same vertical position). The angle is the same too, but with opposite orientation. - Horizontal speeds are the same throughout the trajectory.    - Vertical speeds are the same only at points of equal height.  The vert. comp. shrinks then grows in opposite direction at a const. rate (-g). The resultant velocity vector’s orientation and magnitude changes, but is always tangent. θ

  5. t = 10 t = 5 t = 15 t = 3 t = 17 t = 0 t = 20 Symmetry and Time Over level ground, the time at the peak is half the hang time. Notice the symmetry of times at equal heights relative to the 10 unit mark. The projectile has covered half its range when it has peaked, but only over level ground.

  6. Max height & hang time depend only on initial vertical velocity - Each initial velocity vector below has the a different magnitude (speed) but each object will spend the same time in the air and reach the same max height. - This is because each vector has the same vertical component. -The projectiles will have different ranges, however. The greater the horizontal component of initial velocity, the greater the range.

  7. Here all launch speeds are the same; only the angle varies. 76° 45° 38° Max Range - Over level ground at a constant launch speed, what angle maximizes the range, R ? - First consider some extremes: When θ = 0, R = 0, since the object is on the ground from the moment it’s launched. - When θ = 90°, the object goes straight up and lands right on the launch site, so R = 0 again. The best angle is 45°, smack dab between the extremes.

  8. Range Formula & Max Range at 45° First find the time. - Note that Δy = 0, since the projectile starts and stops at ground level (no change). Δy = v0t + ½ at2. - So, 0 = (v0sinθ)t - ½ at2 - We divide through by t giving us t = 2v0sinθ/a. Then, R = (v0cosθ)t = (v0cosθ)(2v0sinθ /a) = (2v02sinθcosθ) /a. By the trig identity sin2θ = 2sinθcosθ,we getR = (v02sin2θ) /a. Since v0and a are fixed, R is at a max when sin2θ is at a max. When the angle, 2θ, is 90°, the sine function is at its maximum of 1. Therefore, θ = 45°.

  9. http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/ProjectileMotion/jarapplet.htmlhttp://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/ProjectileMotion/jarapplet.html http://www.walter-fendt.de/ph11e/projectile.htm C:\Documents and Settings\hstdilsaver\Local Settings\Temp\phet-projectile-motion\projectile-motion_en.html http://phet.colorado.edu/simulations/sims.php?sim=projectile_motion

  10. Max Range when Δy=0 - When fired from a cliff, or from below ground, a projectile doesn’t attain its max range at 45°. - 45° is only the best angle when a projectile is fired over level ground. - When fired from a cliff, a projectile attains max range with a launch angle less than 45 °(see next slide). - When fired from below ground, a projectile attains max range with a launch angle greater than 45 °.

  11. < 45° 45° If ground were level, the 45° launch would win. Because the < 45° parabola is flatter it eventually overtakes 45 ° parabola. Launch speeds are the same. Range when fired from cliff

  12. 50° 75° 40° 15° Ranges at complementary launch angles - An object fired at angle θ will have the same range as when it’s fired at the same speed at an angle 90° - θ. Reason: R = 2v02sinθ cosθ / a, and the sine of an angle is the cosine of its complement (and vice versa). For example, R at 40° is = 2v02sin40°cos40°/ a = 2v02 sin (90 - 40°)cos(90 - 40°)/a = 2v02cos50°sin50°/ a = R at 50°.

  13. Monkey in a Tree Here’s a classic physics problem: You want to shoot a banana at a monkey up in a tree. Knowing that the monkey will get scared and let go of the branch the instant he hears the sound of the banana gun, how should you aim: a little above, a little below, or right at him? Monkey in a tree web site

  14. Monkey in a tree explanation The reason you should aim right at the monkey even though the monkey lets go right when you pull the trigger is because both the monkey and the banana are in the air for the same amount of time before the collision. So, with respect to where they would have been with no gravity, they fall the same distance. “gravity-free” monkey banana path without gravity monkey w/ gravity path with gravity

  15. 130 m Homerun Example From home plate to the center field wall at a ball park is 130 m. When a batter hits a long drive the ball leaves his bat 1 m off the ground with a velocity of 40 m/s at 28° above the horizontal. The center field wall is 2.6 m high. Does he hit a homerun? 40 m/s 2.6 m 28° 1 m } Let’s first check the range to see if it even has a chance:R = v02sin2θ/a = 402sin56°/9.8 m/s2 = 135.35 m. We need to determine its vertical position when its horizontal position is 130 m. If it’s 1.6 m or more, it’s a homer. Let’s first find the time when the ball is 130 m away (horizontally) from the point where it was hit.

  16. Homerun (cont.) t = d/v = (130 m) / (40cos28°) = 3.68085 s. Let’s see how high up it is at this time: Δy = v0sinθt + ½ at 2. Δy = (40sin28°)(3.68085) - 4.9(3.68085)2 = 2.73 m which is 3.73 m above the ground, out of the reach of a leaping outfielder. Therefore, it’s a home run!