Two Dimensional Motion and Vectors

Two Dimensional Motion and Vectors Projectile Motion

Projectile Motion • Use components to simplify calculations • Avoid vector multiplication • Projectile motion – the motion of objects moving in two dimensions under the influence of gravity • Projectile – an object thrown or launched into the air and subject to gravity

Projectile Motion • Path of a projectile is a parabola if air resistance is disregarded • Any initial horizontal velocity means the object will have horizontal motion • For our purposes, horizontal velocity remains constant • Air resistance would slow it down

Projectile Motion • Projectile motion is free fall with initial horizontal velocity • An object in free fall and a projectile released at the same time, from the same height have the same vertical velocity • Would hit the ground at the same time

Projectile Motion • Vertical motion of a projectile that falls from rest vyf = g * Δt vyf2 = 2 * g * Δy Δy = ½ * g * (Δt)2 • Horizontal Motion of a Projectile vx = vxi = constant Δx = vx * Δt • Use the Pythagorean theorem and inverse tangent functions to find velocity at any point on projectile motion

Projectile Motion The Royal Gorge Bridge in Colorado rises 321m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock’s horizontal displacement is 45.0m. Find the speed at which the rock was kicked.

Projectile Motion Δy = (-321m) Δx = 45.0m g = (-9.81m/s2) vi = vx = ? vx = Δx / Δt Δt=? Δt = √(2 * Δy / g) = √(2(-321) / (-9.81)) =8.04s vx = Δx / Δt =45.0 / 8.04 =5.56m/s y x

Projectiles Launched at an Angle • Must take the angle into consideration in the calculations vyf = vi * (sinθ) + g * Δt vyf2 = vi2 * (sinθ)2 + 2 * g * Δy Δy = vi * (sinθ) * Δt + ½ * g * (Δt)2 vx = vi * (cosθ) = constant Δx = vi * (cosθ) * Δt

Projectiles Launched at an Angle A zookeeper finds an escaped monkey on a pole. While aiming her tranquilizer gun at the monkey, she kneels 10.0m from the pole, which is 5.00m high. The tip of her gun is 1.00m above the ground. At the moment the zookeeper shoots, the monkey drops a banana. The dart travels at 50.0m/s. Will the dart hit the monkey, the banana, or neither one?

Projectiles Launched at an Angle y 4.00m Δx=10.0m g=-9.81m/s2 vi=50.0m/s Δy=4.00m θ=? θ=tan-1(opp / adj) θ=tan-1(Δy / Δx) θ=tan-1(4.00 / 10.0) θ=21.8° Δy=1/2 * g * Δt2 Δt=? Δt=Δx / (vi * cosθ) Δt=10.0 / (50.0cos 21.8) Δt=.215s ? 1.00m x 10.0m

Projectiles Launched at an Angle Δyb = ½ * g * Δt2 = ½(-9.81)(.215) 2 = -.227m Δyd = vi * (sinθ) * Δt + ½ * g * Δt2 = 50.0(sin 21.8)(.215)+ ½(-9.81)(.215) 2 = 3.76m • Banana fell .227m • Dart rises 3.76m • Banana is now {5 - 0.227 = 4.77m} above the ground • Dart is now {1 + 3.76 = 4.76m} above the ground • Answer: Dart hits banana

Two Dimensional Motion and Vectors