1. 3-8 Solving Problems Involving Projectile Motion
Example 3-10: Level horizontal range.
(a) Derive a formula for the horizontal range R of a projectile in terms of its initial speed v0 and angle θ0. The horizontal range is defined as the horizontal distance the projectile travels before returning to its original height (which is typically the ground); that is, y(final) = y0. (b) Suppose one of Napoleon’s cannons had a muzzle speed, v0, of 60.0 m/s. At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?
Figure Caption: Example 3–10. (a) The range R of a projectile; (b) there are generally two angles θ0 that will give the same range. Can you show that if one angle is θ01, the other is θ02 = 90° - θ01 ?
Answer: a. Eliminating t from the kinematics equations, and solving for x when y = 0, we get R = (v02 sin 2θ0)/g.
b. There is a unique solution for sin 2θ0, but two solutions for the angle: θ0 = 30.3° or 59.7°.

2. 3-8 Solving Problems Involving Projectile Motion
Example 3-11: A punt.
Suppose the football in Example 3–7 was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground? Set x0 = 0, y0 = 0.
Figure Caption: Example 3–11: the football leaves the punter’s foot at y = 0, and reaches the ground where y = –1.00 m.
Answer: We can find t from the vertical motion; ignoring the negative solution in the quadratic gives t = 2.53 s. Then we find the horizontal distance to be 40.5 m.

3. 3-8 Solving Problems Involving Projectile Motion
Example 3-12: Rescue helicopter drops supplies.
A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped? (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?
Figure 3-29.
Answer: a. We can find the time the package will take to fall, and then the horizontal distance it will travel: 450 m
b. We know the horizontal distance and the horizontal speed (the speed of the helicopter), so we can find the time the package will take to fall. Substituting gives the initial y velocity: -7.0 m/s (downward)
c. 94 m/s

4. 3-8 Solving Problems Involving Projectile Motion
Projectile motion is parabolic:
Taking the equations for x and y as a function of time, and combining them to eliminate t, we find y as a function of x:
This is the equation for a parabola.

5. 3-8 Solving Problems Involving Projectile Motion
Examples of projectile motion. Notice the effects of air resistance.
Figure Caption: Examples of projectile motion—sparks (small hot glowing pieces of metal), water, and fireworks. The parabolic path characteristic of projectile motion is affected by air resistance.

6. 3-9 Relative Velocity
We have already considered relative speed in one dimension; it is similar in two dimensions except that we must add and subtract velocities as vectors.
Each velocity is labeled first with the object, and second with the reference frame in which it has this velocity.

7. 3-9 Relative Velocity
Here, vWS is the velocity of the water in the shore frame, vBS is the velocity of the boat in the shore frame, and vBW is the velocity of the boat in the water frame.
The relationship between the three velocities is:
Figure Caption: To move directly across the river, the boat must head upstream at an angle θ. Velocity vectors are shown as green arrows:
vBS = velocity of Boat with respect to the Shore,
vBW = velocity of Boat with respect to the Water,
vWS = velocity of the Water with respect to the Shore (river current).

8. 3-9 Relative Velocity Example 3-14: Heading upstream.
A boat’s speed in still water is vBW = 1.85 m/s. If the boat is to travel directly across a river whose current has speed vWS = 1.20 m/s, at what upstream angle must the boat head?
Figure 3-33.
Answer: 40.4° (simple geometry)

9. 3-9 Relative Velocity Example 3-15: Heading across the river.
The same boat (vBW = 1.85 m/s) now heads directly across the river whose current is still 1.20 m/s. (a) What is the velocity (magnitude and direction) of the boat relative to the shore? (b) If the river is 110 m wide, how long will it take to cross and how far downstream will the boat be then?
Figure Caption: Example 3–15. A boat heading directly across a river whose current moves at 1.20 m/s.
Answer: a m/s at an angle of 33.0° downstream.
b s and 71.4 m

10. 3-9 Relative Velocity Example 3-16: Car velocities at 90°.
Two automobiles approach a street corner at right angles to each other with the same speed of 40.0 km/h (= 11.1 m/s), as shown. What is the relative velocity of one car with respect to the other? That is, determine the velocity of car 1 as seen by car 2.
Figure 3-35.
Answer: 15.7 m/s

11. Review Questions

12. Which inequality does the vector sum of the two vectors A and B necessarily satisfy?
The magnitude of A+B is greater than or equal to the magnitude of A and greater than or equal to the magnitude of B.
The magnitude of A+B is greater than or equal to the magnitude of A plus the magnitude of B.
The magnitude of A+B is less than or equal to the magnitude of A and less than or equal to the magnitude of B.
The magnitude of A+B is less than or equal to the magnitude of A plus the magnitude of B.

13. A projectile is launched with an initial speed v
A projectile is launched with an initial speed v. Which of the following statements is necessarily true at any point along its trajectory above the initial launch position?
The speed of the projectile is greater than the launch speed.
The speed of the projectile is equal to the launch speed.
The speed of the projectile is less than the launch speed.
The angle of the velocity is steeper than the angle of the launch velocity.

14. What is/are the condition(s) of the motion of a projectile at its highest point along its trajectory?
The speed and acceleration are zero.
The horizontal velocity and the acceleration are zero.
The vertical velocity and the acceleration are zero.
The vertical velocity is zero.

15. In order to take off, a plane must have an airspeed of 60 mi/h
In order to take off, a plane must have an airspeed of 60 mi/h. What is the required groundspeed if the plane is taking off with a headwind of 15 mi/h?
75 mi/h
60 mi/h
45 mi/h

16. Homework Ch .3
Read Ch. 3.
Ch. 3 Problems #’s 11, 17, 23, 29, 35, 39, 41, 43, 51, 57, 63