Physics pre-AP

Equations of motion :

We assume NO AIR RESISTANCE! (Welcome to “Physicsland”), therefore… The path of a projectile is a parabola. Horizontal motion is constant velocity. Vertical motion is in “free-fall”. Vertical velocity at the top of the path is zero Time is the same for both horizontal and vertical motions. The horizontal and vertical motions are independent. Concepts concerning Projectile Motion:

horizontal or “x” – direction vertical or “y” – direction Remember that for projectiles, the horizontal and vertical motions must be separated and analyzed independently. Remember that “a x ” is zero and “a y ” is acceleration due to gravity “g = 9.81 m/s 2 ”

A cannon ball is shot horizontally from a cliff. vxvx height dydy Range, d x What do we know? For all projectiles… Hint: You should always list your known values at the beginning of any problem and assign those values variables. vxvx

vxvx height dydy Range, d x Remember to keep the horizontal and vertical motions separate. Time is the same for both directions. The time to fall (flight time) depends on the height from which the projectile falls. Knowns:

A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time, t = ? Find range, d x = ? Find final velocity, v f = ? V x = 5 m/s height d y =35 m Range, d x Let’s begin by finding time. Knowns: Givens:

V x = 5 m/s height d y =35 m Range, d x Start with the distance equation… Now solve for t…

V x = 5 m/s height d y =35 m Range, d x Now that we know time, let’s find d x (the range). Remember that horizontal motion is constant so there is no acceleration. Let’s use the distance formula again. This time in the x – direction… 0 Notice: the distance in the horizontal direction is the horizontal velocity times the time. d x = v x · t

V x = 5 m/s height d y =35 m Range, d x Final velocity is a vector quantity so we must state our answer as a magnitude (speed that the projectile strikes the ground) and direction (angle the projectile strikes the ground). Also remember horizontal velocity is constant, therefore the projectile will never strike the ground exactly at 90°. That means we need to look at the horizontal (x) and vertical (y) components that make up the final velocity. Calculated: VfVf Determining the velocity at impact: there are two components to the final velocity vector. The constant v x and the accelerated v y.

V x = 5 m/s height d y =35 m Range, d x Calculated: v impact V fx = 5 m/s (constant) V fy θ θ So, we have the x-component already due to the fact that horizontal velocity is constant. Before we can find v impact, we must find the vertical component, v fy. Calculating v fy : To find v fy, remember that vertical motion is in “free- fall” so it is accelerated by gravity from zero to some value just before it hits the ground.

V x = 5 m/s height d y =35 m Range, d x Now we know both the x- and y- components of the final velocity vector. We need to put them together for magnitude and direction of final velocity. θ Putting it together to calculate v f : Final Velocity = 26.7 m/s, 79.2° Putting everything together:

vivi θ Since the initial velocity represents motion in both the horizontal (x) and vertical (y) directions at the same time, we cannot use it in any of our equations. Remember, the most important thing about projectiles is that we must treat the horizontal (x) and vertical (y) completely separate from each other. So…we need to separate “v i ” into its x- and y- components. We will use the method we used for vectors. v iy θ v ix vivi Now that we have the components of the initial velocity, we will use only those for calculations. v iy

vivi θ A projectile’s path is a parabola, ALWAYS. That means if a projectile is launched and lands at the same height, there will be symmetry. The angle of launch and angle of landing will be equal. The initial velocity and the final velocity will be the same magnitude. Also, that means the components will be the same. v iy v ix vfvf θ v fx v fy Since the horizontal motion is always at constant velocity… Since the vertical motion is the same as a ball that is thrown straight up or dropped straight down (in free-fall), the y- components are equal and opposite. vfvf θ v fx v fy

vivi θ Step 1: List known values! Draw and label picture. Knowns (for all projectiles): V y top = 0 V iy VxVx d ymax =height Range, dx

vivi θ VxVx V y top = 0 VyVy d ymax =height Range, dx Step 2: Divide initial velocity into horizontal (x) and vertical (y) components. Step 3: Find flight time if possible. Use vertical motion. Keeping the horizontal and vertical motions separate!

vivi θ Almost every projectile problem can be solved by starting with the displacement equation to solve for time. In this case… Now solve for time. Yes, it is a quadratic equation! This will be the time for the entire flight. NOTE: If you want to find maximum height you will only use half the time. If you want to find range, use the total time. Finding time – Method 1: Since the initial and final vertical positions are both the same, vertical displacement d y = 0. V y top = 0 VyVy d ymax =height Range, dx Note: There are three ways to find time for this problem. You may use any of them you wish.

V i = 30 m/s 60 ° Step 1: List known values! Draw and label picture. Knowns (for all projectiles): V y top = 0 V iy VxVx d ymax =height Range, dx A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. Given values:

V i = 30 m/s 60 ° Step 2: Divide initial velocity into x- & y- components. Knowns (for all projectiles): V y top = 0 V iy VxVx d ymax =height Range, dx A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. Given values: We can add these to what we know. WE WILL NOT USE THE 30 m/s again in this problem because it is not purely an x- or y- value.

Finding time – Remember that d y = 0 because the projectile is starting and ending at the same level (y- position). So, using the known and given values for this problem and the components we calculated, we can solve for time. V y top = 0 v y =26 m/s d ymax =height Range, dx v i = 30 m/s 60° A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. v ix =15 m/s

vivi θ V y top = 0 VyVy VxVx d ymax =height Range, dx vfvf θ v fx v fy Now that I know time, I can add it to my list of known, given, and calculated values. To review… Knowns (for all projectiles):Givens: Calculated values: A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.

vivi θ V y top = 0 d ymax =height Range, dx Now that we know time, let’s calculate horizontal distance. Remember that horizontal acceleration is zero. vfvf θ v y =26 m/s v y =-26 m/s A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. v x =15 m/s 0

vivi θ V y top = 0 d ymax =height Range, dx To find maximum height for this problem, remember that v ytop = 0 So: vfvf θ v y =26 m/s v y =-26 m/s A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height. v x =15 m/s

Vertical displacement is not zero. Consider the launch point as the zero height (reference point) If you know the range: use range to find the time to that position. Δx = v x t Use this time in the distance equation to find the height at that position. vivi θ dydy dxdx NOTE: The highest point of the projectile DOES NOT occur at the half- way point of the flight. BE CAREFUL! Projectiles that do not land at the same height:

vivi θ -d y dxdx Again, if the range is known use the method in the previous slide. If the range is not known, use quadratic formula to find the time to the landing. Then use this time to find the range. NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL!