Angled Launch Physics

Projectile: Any object that is in free fall. Examples: pumpkins sky divers arrows rocks vomit etc…

Angled Launch L  Launch velocity v0 Vertical launch velocity v0y = v0sin Horizontal launch velocity y- dimension: v0y fights off gravity and gives the projectile air time. x- dimension: v0x keeps the projectile moving forward at a constant speed until it strikes something. v0x = v0cos

Let’s analyze a soccer ball that has been kicked at an angle. 20 m/s 20 m/s 9.8 m/s 19.6 m/s 20 m/s 20 m/s -9.8 m/s 29.4 m/s 20 m/s 20 m/s -19.6 m/s y – dimension loses 9.8 m/s every second x – dimension continues unaffected Notice the launch velocity = impact velocity -29.4 m/s 20 m/s

SYMBOLIC KINEMATICS END ∆x = v0 = v = a = t = Y – DIMENSION: END ∆x = v0 = v = a = t = TOP ∆x = v0 = v = a = t = 0 m height Lsin Lsin - Lsin 0 m/s - 9.8 m/s2 - 9.8 m/s2 ttotal tup = ½ ttotal

SYMBOLIC KINEMATICS X – DIMENSION: ∆x = v = t = distance Lcos ttotal

NUMERIC EXAMPLE: END ∆x = v0 = v = a = t = Bob launches a rocket moving 150 m/s at an angle of 35 degrees. How high will it go? Where will it land? END ∆x = v0 = v = a = t = Use Equation 1 to solve for time: 0 m Lsin = 150 sin (35) = 86.0 v = at + v0 - Lsin = -150 sin (35) = -86.0 t = v - v0 a = -86 – 86 -9.8 - 9.8 m/s2 = 17.55855 ttotal = need

NUMERIC EXAMPLE: TOP Use Equation 3 to solve for ∆x : ∆x = v0 = want o Bob launches a rocket moving 150 m/s at an angle of 35 degrees. How high will it go? Where will it land? TOP ∆x = v0 = v = a = t = Use Equation 3 to solve for ∆x : want o Lsin = 150 sin (35) = 86.0 ∆x = ½ (v0 +v)t 0 m/s ∆x = ½ (v0)t = ½ (86)(8.779235) - 9.8 m/s2 tup = 17.55855/2 = 8.779235 = 378 m

NUMERIC EXAMPLE: ∆x = v = want t = Lcos = 150 cos (35) = 122.8728 Bob launches a rocket moving 150 m/s at an angle of 35 degrees. How high will it go? Where will it land? ∆x = v = t = want Lcos = 150 cos (35) = 122.8728 17.55855 Use the velocity equation to solve for ∆x : v = ∆x ∆t ∆x = v∆t = (122.8728)(17.55855) = 2157 m

NUMERIC EXAMPLE: How high and far will Bob’s rocket go if he changes the angle to 45 degrees?

NUMERIC EXAMPLE: How high and far will Bob’s rocket go if he changes the angle to 55 degrees?

MAXIMUM DISTANCE & HEIGHT Let’s make a table of what we learned. It turns out there are some useful patterns we have observed. ANGLE HEIGHT DISTANCE 35 45 55 378 m 574 m 770 m 2157 m 2296 m

SAME IMPACT SITE: MAX DISTANCE: MAX HEIGHT: If two launch angles are complementary (add up to 90 degrees) they will both land at the same location. (One goes high, one goes low, but they both hit the same target) If you want to obtain the maximum distance, launch with a 45 degree angle. If you want to obtain the maximum height, launch with a 90 degree angle.

Advanced Angled Launch Physics

Advanced Angled Launch What if you are looking for a point on the trajectory that is not located at the maximum height or distance? IF YOU KNOW THE TIME: Plug it in to both the x and y dimensions.

EXAMPLE: X ∆x = v = t = Y ∆x = v0 = v = a = t = want want 13 cos (40) An angry bird is launched 13 m/s at an upward angle of 40 degrees. Where will the bird be half a second later? X ∆x = v = t = Y ∆x = v0 = v = a = t = want want 13 m/s 13 cos (40) 13 sin (40) 40 degrees 0.5 s don’t care - 9.8 m/s2 Image from: www.iconarchive.com v = ∆x ∆t ∆x = v∆t 0.5 s = (13 cos (40))(0.5) ∆x = ½ at2 + v0t ∆x = ½ (-9.8)(0.5)2 + (13 sin (40)(0.5) = 4.98 m = 2.95 m

Advanced Angled Launch IF YOU KNOW THE HORIZONTAL POSITION: Use the x-dimension to solve for time and then plug it in.

EXAMPLE: X ∆x = v = t = Y ∆x = v0 = v = a = t = 1.2 m want 8 cos (32) Bob uses his spoon to catapult some mashed potatoes at the wall. The wall is 1.2 meters away from his chair. The potatoes leave his spoon moving 8 m/s at an upward angle of 32 degrees. Where on the wall will they strike? X ∆x = v = t = Y ∆x = v0 = v = a = t = 1.2 m want 8 m/s 8 cos (32) 8 sin (32) need don’t care 32 degrees - 9.8 m/s2 v = ∆x ∆t ∆t = ∆x/v 0.177 s = 1.2/(8 cos (32)) ∆x = ½ at2 + v0t ∆x = ½ (-9.8)(0.177)2 + (8 sin (32)(0.177) = 0.177 s = 0.597 m above the spoon

Advanced Angled Launch IF YOU WANT TO KNOW WHETHER THE PROJECTILE WAS MOVING UPWARDS OR DOWNWARDS: Calculate the time UP and compare it with the current time.

EXAMPLE: Y ∆x = v0 = max height v = at + v0 v = a = 13 sin (40) Was the angry bird from example 1 still moving up or was it coming back down? Y ∆x = v0 = v = a = t = max height v = at + v0 13 sin (40) t = - v0 / a 0 m/s = - (13 sin (40)) / (-9.8) - 9.8 m/s2 = 0.853 s time up The bird can go up for 0.853 seconds. It is currently only 0.5 seconds and therefore the bird is going UP.

EXAMPLE: Y ∆x = v0 = max height v = at + v0 v = a = 8 sin (32) Were the potatoes from example 2 still moving up or were they coming back down? Y ∆x = v0 = v = a = t = max height v = at + v0 8 sin (32) t = - v0 / a 0 m/s = - (8 sin (32)) / (-9.8) - 9.8 m/s2 = 0.433 s time up The potatoes can go up for 0.433 seconds. It is currently only 0.177 seconds and therefore the potatoes are still going UP.