# Projectiles at an Angle

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Projectiles at an Angle
Hardest thing yet.

A progression: Freefall Review:
A ball drops for 3 s. How far does it go? How fast is it going after 3 s?

Next… B) Throwing upward:
A ball is thrown upward with an initial velocity of 30 m/s. How far does the ball travel? How long before it hits the ground? Divide the motion into two halves- up and down Symmetry: timeup = timedown viup = -vfdown dup = - ddown

Next… x y vi = vf = a = d = time = C) Horizontal Projectiles:
A balls rolls off a table with a velocity of 10 m/s and lands 30 m away from the table. How long did it take the ball to hit the ground? x y vi = vf = a = d = time = Mythbusters

The hardest… D) Projectiles launched at an angle:
A ball is launched with a velocity of 100 m/s at 30º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball? Height θ Range

Vi = 100m/s Height θ = 30º Range Our projectile is no longer launched horizontally so it no longer has an viy= 0 m/s The vi is divided between the x and y directions- we need to do something called “Resolving the x and y components”

Let’s focus on the angle:

Trigonometry- finding the side or angles of a triangle
Vi = 100 m/s h Viy o θ = 30º θ Vix a SOHCAHTOA

Let’s find vix (a) knowing vi (h):
Vi = 100 m/s h Viy o θ = 30º θ Vix a

Let’s find viy (o) knowing vi (h):
Vi = 100 m/s h Viy o θ = 30º θ Vix a

Let’s use these new formulas to find the vix and viy for our example:

Stop and Check: 100 m/s 50 m/s θ 86.6 m/s

Now we can take this information and put it into a new form of our table:
vi = θ = x y vf = a = d = ½ t = t =

Revisit our problem: x y vi = 86.6 m/s vf = 86.6 m/s a = 0 m/s d =
A ball is launched with a velocity of 100 m/s at 30º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball? vi = 100 m/s θ = 30º x y vi = 86.6 m/s vf = 86.6 m/s a = 0 m/s d = vi = 50 m/s vf = -50 m/s a = -10 m/s2 ½ t = t =

1. Find the maximum height (dy):
Vi = 100m/s Height θ = 30º Range HEIGHT WILL LEAD YOU TO TIME

2. Finding time: Vtop = Vi = 100m/s θ = 30º Vf = Range
Now this becomes like a horizontal projectile problem. Find time. Remember tup = tdown

3. Finding the range (dx):

4. Finding components using a ruler and a protractor:

Which launch angle gives the greatest range?
The cannonball launched at a 45-degree angle had the greatest range. The cannonball launched at a 60-degree angle had the highest peak height before falling. The cannonball launched at the 30-degree angle reached the ground first.

Practice Problem: A golf ball is hit with a velocity of 20 m/s at 45 above the horizon. Find the time it takes the golf ball to travel, the range of the golf ball, and the maximum height the golf ball reaches.

A player kicks a football from ground level at 27
A player kicks a football from ground level at 27.0 m/s at an angle of 30.0 degrees above the horizontal. Find: its “hang time” (time that the ball is in the air), the distance the ball travels before it hits the ground, and its maximum height.

A cannon is fired. The cannonball reaches a maximum height of 22. 5 m
A cannon is fired. The cannonball reaches a maximum height of 22.5 m. The range of the cannonball is 89.9 m. At what angle was the cannon fired from? Height = 22.5 m θ = ? Range = 89.9 m