Projectile Motion Practice Horizontal Launch This topic can be found on pp. 96-98 of your book. “Waterfall” (1961) MC Escher

Rocky Rhodes is travelling at 35 m/s on a level road when he swerves off the road and over a cliff. If it takes him 5.3 s to hit the ground, a) how far away from the cliff base did Rocky hit the ground and b) how high is the cliff? Practice 1

Practice 1 Let’s first draw a picture. Δx = ? Rocky Rhodes is travelling at 35 m/s on a level road when he swerves off the road and over a cliff. If it takes him 5.3 s to hit the ground, a) how far away from the cliff base did Rocky hit the ground and b) how high is the cliff? Practice 1 Let’s first draw a picture. Δx = ?

Practice 1 x-direction y-direction Δx = ? Δy = ? t = 5.3 s v = 35 m/s Rocky Rhodes is travelling at 35 m/s on a level road when he swerves off the road and over a cliff. If it takes him 5.3 s to hit the ground, a) how far away from the cliff base did Rocky hit the ground and b) how high is the cliff? Practice 1 Next, let’s use the GUESS method to solve part (a) of this problem Givens and Unknowns x-direction y-direction Δx = ? Δy = ? t = 5.3 s v = 35 m/s vi = 0 m/s vf = ? a = -9.81 m/s2 Δx = ?

Practice 1 x-direction y-direction Δx = ? Δy = ? t = 5.3 s v = 35 m/s vi = 0 m/s vf = ? a = -9.81 m/s2 Because there is constant velocity in the x-direction, we can use the following equation to solve for Δx. 𝑣= Δ𝑥 𝑡 Δx = ?

Practice 1 x-direction y-direction Δx = ? Δy = ? t = 5.3 s v = 35 m/s vi = 0 m/s vf = ? a = -9.81 m/s2 𝑣= Δ𝑥 𝑡 35 𝑚/𝑠= Δ𝑥 5.3 𝑠 Δx = 190 m (185.5 m) Δx = ?

Practice 1 x-direction y-direction Δx = ? Δy = ? t = 5.3 s v = 35 m/s vi = 0 m/s vf = ? a = -9.81 m/s2 Now let’s solve part (b). Notice that we have three variables in the y-direction. We can use kinematics to solve for Δy. ∆𝑦= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 Δx = ?

∆𝑦=(0𝑚/𝑠)(5.3𝑠)+ 1 2 (−9.81𝑚/ 𝑠 2 )(5.3𝑠) 2 Practice 1 x-direction y-direction Δx = ? Δy = ? t = 5.3 s v = 35 m/s vi = 0 m/s vf = ? a = -9.81 m/s2 ∆𝑦= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 ∆𝑦=(0𝑚/𝑠)(5.3𝑠)+ 1 2 (−9.81𝑚/ 𝑠 2 )(5.3𝑠) 2 Δy = -140 m (-137.6 m) Δx = ?

Rino the hamster is rolling in his hamster ball across a table that is 0.92 m tall. He rolls off the table and lands 0.34 m away from the base of the table. How fast was Rino travelling when he rolled off the table? Practice 2

Practice 2 x-direction y-direction Rino the hamster is rolling in his hamster ball across a table that is 0.92 m tall. He rolls off the table and lands 0.34 m away from the base of the table. How fast was Rino travelling when he rolled off the table? Practice 2 Write a chart and list your knowns and unknowns. x-direction y-direction

Rino the hamster is rolling in his hamster ball across a table that is 0.92 m tall. He rolls off the table and lands 0.34 m away from the base of the table. How fast was Rino travelling when he rolled off the table? Practice 2 Write a chart and list your knowns and unknowns. We don’t have enough information in the x-direction right now to solve this problem, but we do have enough in the y-direction to solve for t. Then we can solve for vx. x-direction y-direction Δx = 0.34 m Δy = -0.92 m t = ? v = ? vi = 0 m/s vf = ? a = -9.81 m/s2

Practice 2 ∆𝑦= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 −0.92𝑚 =(0𝑚/𝑠)(𝑡)+ 1 2 (−9.81𝑚/ 𝑠 2 )𝑡 2 t = 0.43 s x-direction y-direction Δx = 0.34 m Δy = -0.92 m t = ? v = ? vi = 0 m/s vf = ? a = -9.81 m/s2

Practice 2 ∆𝑦= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 −0.92𝑚 =(0𝑚/𝑠)(𝑡)+ 1 2 (−9.81𝑚/ 𝑠 2 )𝑡 2 t = 0.43 s x-direction y-direction Δx = 0.34 m Δy = -0.92 m t = ? t = 0.43 s v = ? vi = 0 m/s vf = ? a = -9.81 m/s2

Practice 2 ∆𝑦= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 −0.92𝑚 =(0𝑚/𝑠)(𝑡)+ 1 2 (−9.81𝑚/ 𝑠 2 )𝑡 2 t = 0.43 s x-direction y-direction Δx = 0.34 m Δy = -0.92 m t = 0.43 s v = ? vi = 0 m/s vf = ? a = -9.81 m/s2 Now we have enough information in the x-direction to solve for vx.

Practice 2 𝑣 𝑥 = ∆𝑥 𝑡 𝑣 𝑥 = 0.34 𝑚 0.43 𝑠 vx = 0.79 m/s x-direction y-direction Δx = 0.34 m Δy = -0.92 m t = 0.43 s v = ? vi = 0 m/s vf = ? a = -9.81 m/s2

Practice 3 A lemming runs towards a 32.7 m cliff at 5.30 m/s and throws itself off the edge. What is the velocity (both magnitude and direction) when it lands in the sea below? Click here to see the famous Disney nature “documentary” that shows this peculiar behavior of lemmings. It was later discovered that the whole event was faked and that lemmings do not commit mass suicide. Click here for info.

Practice 3 First, let’s draw a picture. vx = 5.30 m/s A lemming runs towards a 32.7 m cliff at 5.30 m/s and throws itself off the edge. What is the velocity (both magnitude and direction) when it lands in the sea below? Δy = 32.7 m vf = ?

Practice 3 x-direction y-direction Δx = ? Δy = -32.7 m t = ? Now let’s create a table and organize what we know and don’t know. A lemming runs towards a 32.7 m cliff at 5.30 m/s and throws itself off the edge. What is the velocity (both magnitude and direction) when it lands in the sea below? x-direction y-direction Δx = ? Δy = -32.7 m t = ? vx = 5.30 m/s vi = 0 m/s vf = ? a = -9.81 m/s2

Practice 3 x-direction y-direction Δx = ? Δy = -32.7 m t = ? In order to solve for the final velocity, we need to know both the x- and y-components of the final velocity. We already know the x-component, but we’ll need to solve for the y-component. A lemming runs towards a 32.7 m cliff at 5.30 m/s and throws itself off the edge. What is the velocity (both magnitude and direction) when it lands in the sea below? x-direction y-direction Δx = ? Δy = -32.7 m t = ? vx = 5.30 m/s vi = 0 m/s vf = ? a = -9.81 m/s2

vf in the y-direction = -25.3 m/s Practice 3 Solving for vf in the y-direction: 𝑣 𝑓 2 = 𝑣 𝑖 2 +2𝑎∆𝑦 𝑣 𝑓 2 = (0 𝑚 𝑠 ) 2 +2(−9.8 𝑚 𝑠 2 )(−32.7 𝑚) vf in the y-direction = -25.3 m/s A lemming runs towards a 32.7 m cliff at 5.30 m/s and throws itself off the edge. What is the velocity (both magnitude and direction) when it lands in the sea below? x-direction y-direction Δx = ? Δy = -32.7 m t = ? vx = 5.30 m/s vi = 0 m/s vf = ? a = -9.81 m/s2

vf in the y-direction = -25.3 m/s Practice 3 Solving for vf in the y-direction: 𝑣 𝑓 2 = 𝑣 𝑖 2 +2𝑎∆𝑦 𝑣 𝑓 2 = (0 𝑚 𝑠 ) 2 +2(−9.8 𝑚 𝑠 2 )(−32.7 𝑚) vf in the y-direction = -25.3 m/s A lemming runs towards a 32.7 m cliff at 5.30 m/s and throws itself off the edge. What is the velocity (both magnitude and direction) when it lands in the sea below? x-direction y-direction Δx = ? Δy = -32.7 m t = ? vx = 5.30 m/s vi = 0 m/s vf = -25.3 m/s a = -9.81 m/s2

Practice 3 x-direction y-direction Δx = ? Δy = -32.7 m t = ? Now that we know both the x- and y-components of the final velocity, we can solve for the resultant vector. x-direction y-direction Δx = ? Δy = -32.7 m t = ? vx = 5.30 m/s vi = 0 m/s vf = -25.3 m/s a = -9.81 m/s2 θ vf = ? -25.3 m/s 5.30 m/s

vf = 25.8 m/s @ 11.8o from the cliff side Practice 3 Now that we know both the x- and y-components of the final velocity, we can solve for the resultant vector. Using the Pythagorean theorem: 𝑣 𝑓 2 = 𝑣 𝑥 2 + 𝑣 𝑦 2 vf2 = (5.30 m/s)2 + (-25.3 m/s)2 vf = 25.8 m/s Using trig, we can get the angle. 𝑡𝑎𝑛𝜃= 𝑜𝑝𝑝 𝑎𝑑𝑗 𝑡𝑎𝑛𝜃= 5.30 𝑚/𝑠 25.3 𝑚/𝑠 θ = 11.8o from the cliff side Final answer: vf = 25.8 m/s @ 11.8o from the cliff side θ vf = ? -25.3 m/s 5.30 m/s

y-direction equations Projectile Motion Equations Summary x-direction equation y-direction equations 𝑣= ∆𝑥 ∆𝑡 𝑣 𝑓 = 𝑣 𝑖 +𝑎∆𝑡 ∆𝑥= 1 2 (𝑣 𝑖 + 𝑣 𝑓 )∆𝑡 ∆𝑥= 𝑣 𝑖 ∆𝑡+ 1 2 𝑎 ∆𝑡 2 ∆𝑥= 𝑣 𝑓 ∆𝑡− 1 2 𝑎 ∆𝑡 2 𝑣 𝑓 2= 𝑣 𝑖 2+2𝑎∆𝑥

Important Points vx is always constant in projectile motion constant!!! Acceleration is only due to gravity (down) and is always the same value: -9.81 m/s2 If a projectile if fired horizontally, viy = 0 m/s. Time is the only variable that must be the same in both the x and y directions and is therefore the link between the two directions.