Projectile problems

A person launches a ball from 1 A person launches a ball from 1.8 m and it lands over a 10m high fence at 35m/s at angle of 40 degrees. What is the launch angle? How long is the ball in the air? How far did it travel?

Resolve final velocity Vx = V cos Θ = 35 cos 40 = 26.8 m/s Vy = V sin Θ = 35 sin 40 = -22.5 m/s (Ball is heading downward) Vx = Vox (No acceleration in the x-direction)

Final initial velocity in the Y- direction Vy2 = V0y2 + 2(a)(Δy) Δy= 10- 1.8= 8.2m -22.52= V0y2 + 2 (-9.8) (8.2) V0y = 25.8 m/s

Find Launch angle Tan-1 = V0y/V0x Launch angle = 43.9°

Time in the air Vy = V0y + at -22.5 = 25.8 + (-9.8) t t = 4.92 sec

Range X = Voxt X = 26.8 (4.92) 131.9 m

A ball rolls of a house at 6 m/s A ball rolls of a house at 6 m/s. The roof slopes at 28° and is 8 m above the ground. How long is the ball in the air? Where will it hit?

Time in air (2 methods) First method Second method Quadratic formula -8 = -6 sin 28 + ½ (-9.8) t2 -4.9t2 +(-2.82)t + 8 = 0 t= -1.59 or 1.02 sec Second method Vy2 = Voy2 + 2 (a) (Δy) Vy= -12.8 m/s (Ball is going down) Vy= Voy + at -12.8 = -2.82 + (-9.8) t t = 1.02 sec

Range from roof’s edge X = Voxt + ½ at2 X = V cos Θ t X = 6 cos 28 (1.02) X= 5.4 m

Relative Motion

What is Relative Motion Strictly speaking…all motion is relative to something. Usually that something is a reference point that is assumed to be at rest (i.e. the earth). Motion can be relative to anything…even another moving object. Relative motion problems involve solving problems with multiple moving objects which may or may not have motion relative to the same reference point. In fact, you may be given motion information relative to each other.

Notation for Relative Motion We use a combination of subscripts to indicate what the quantity represents and what it is relative to. For example, “va/b” would indicate the velocity of “object a” with respect to “object b”. Object b in this example is the reference point. Note: The “reference point” object is assumed to be at rest.

Example Problem A plane flies due north with an airspeed of 50 m/s, while the wind is blowing 15 m/s due East. What is the speed and direction of the plane with respect to the earth? What do we know? “Airspeed” means the speed of the plane with respect to the air. “wind blowing” refers to speed of the air with respect to the earth. What are we looking for? “speed” of the plane with respect to the earth. We know that the speed and heading of the plane will be affected by both it’s airspeed and the wind velocity, so… just add the vectors.

Example Problem (cont.) So, we are adding these vectors…what does it look like? Draw a diagram,of the vectors tip to tail! Solve it! This one is fairly simple to solve once it is set up…but, that can be the tricky part. Let’s look at how the vector equation is put together and how it leads us to this drawing. N θ

How to write the vector addition formula middle same first last Note: We can use the subscripts to properly line up the equation. We can then rearrange that equation to solve for any of the vectors. Always draw the vector diagram, then you can solve for any of the vector quantities that might be missing using components or even the law of sines.

Displacement is relative too! Other quantities can be solved for in this way, including displacement. Remember that d=vt and so it is possible to see a problem that may give you some displacement information and other velocity information but not enough of either to answer the question directly When solving these, be very careful that all the quantities on your diagram and in your vector formula are alike (i.e. all velocity or all displacement). Do not mix them!

1 –D Relative motion If car A is moving 5m/s East and car B, is moving 2 m/s West, what is car A’s speed relative to car B. 5 m/s Car A Car B 2 m/s So, we want to know…if we are sitting in car B, how fast does car A seem to be approaching us? Common sense tells us that Car A is coming at us at a rate of 7 m/s. How do we reconcile that with the formulas?

1- D and the vector addition formula Let’s start with defining the reference frame for the values given. Both cars have speeds given with respect to the earth. Va/e =5 m/s Car A Car B Vb/e = -2 m/s We are looking for the velocity of A with respect to B, so va/b = ? If we set up the formula using the subscript alignment to tell us what to add, we get… Then we need to solve for va/b . So…

A swimmer heads directly across a river, swimming at 1 A swimmer heads directly across a river, swimming at 1.6 m/s relative to still water. They arrive at a point 40 m downstream from the point directly across the river, which is 80m wide. 1) What is the speed of the river current? 2) What is the swimmer’s speed relative to the shore? 3) What direction should the swimmer head so as to arrive at the point directly opposite of starting point?

Speed of the river Create x-y system Swimmer goes across river on y-axis River flows on the x-axis Vy = d/t 1.6 = 80/t ; t= 50 seconds for total trip If time to travel y is 50 seconds then equal time used to travel in x direction since motions happen simultaneously Vx = x/t Vx = 40/50 = .8 m/s

Swimmer’s speed relative to the shore Vs/e= Vs/w + Vw/e Vector addition of perpendicular vectors Vs/e2 = Vs/w2 + Vw/e2 𝑎 2 + 𝑏 2 = 𝑐 2 1.62 + .82 = Vs/e2 Vs/e = 1.79 m/s

Swimmer’s direction to end straight across river .8 1.6 Vs/e = 1.6 cos Θ = .8 Θ = 60° from the shore upstream