I need help with my AP Physics homework (Projectile Motion)?

Question

Can anyone describe to me how to do projectile motion?

I have some problems that I need help with:

1. A fire hose ejects a stream of water at 35° above the horizontal. the water leaves the nozzle with a speed of 25 m/s. Assuming that the water behaves like a projectile, how far from the building should the fire hose be located to hit the highest possible fire?

2. A horizontal rifle is fired at a bullseye. The muzzle speed of the bullet is 670 m/s. The barrel is pointed at the center of the bull's-eye, but the bullet strikes the target .025 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?

3. An Olympic long jumper leaves the ground at an angle of 23° and travels through the air for a horizontal distance of 8.7 m before landing. What is the takeoff speed of the jumper?


I would really appreciate your help I have a test on this tomorrow! :(

Answer 1

1. Consider the vertical component of velocity working against gravity

v=sin(35)*25

using conservation of energy to find the apogee:

1/2*m*v^2=m*g*h
or
h=v^2/(2*g)

now calculate the flight time using average velocity and constant acceleration:

h=1/2*v*t
t=2*h/v
from above:
t=2*v^2/(2*g*v)
t=v/g
or
t=sin(35)*25/9.81

once you have t, the distance from the building can be calculated using the horizontal velocity component
d=cos(35)*25*t
d=cos(35)*25*sin(35)*25/9.81
d=29.93m

2. The bullet will descend from the muzzel to the target under the influence of gravity.

Using flight time of the bullet as t:
d=1/2*g*t^2
from the problem statement
d=.025
t=sqrt(2*.025/9.81)
=.0714 s
Given that and the muzzel velocity:
horizontal distance=
670*t
=47.8 m


3. Again, look at the flight time. Since the jumper will reach apogee and return
g*t^2=h
v*sin(23)*t=h
t=v*sin(23)/g

The flight time in the horizontal
v*cos(23)*t=8.7
v^2*cos(23)*sin(23)/g=8.7
v=sqrt(9.81*8.7/(cos(23)*sin(23)))
=15.4 m/s

j

Answer 2

Well, there are others who will answer your question, perhaps i'll try to help u in understanding projectile motions.

As you may already know, when an object is projected at an angle, there is always a horizontal component (VcosӨ) and a vertical component (VsinӨ).

Since the force of gravity acts downwards, it can only affect the up and down motion, while the speed of the sideways motion remain constant, neglecting air resistance/drag/any resistance.
1 thing to note, the vertical and horizontal components act INDEPENDENTLY.

Movement of projectiles can be solved (generally) as follows:
1. First find the separate vertical and horizontal velocities by simply computing the right values in the expression above.

2. Calculate the time of flight, t (i.e the time the object takes to reach its highest point) by the formula Vf = Vi + at, which is simplified to v = gt, where v is the vertical component of the velocity.
If you do not understand how this is the way it is, view it as if u're throwing a ball vertically up. The time it takes to reach its highest point where its velocity is zero (the instant b4 it drops) is given by Vf = Vi + at, where, Vf = 0, Vi is the speed u throw it, a = g and is negative.

3. Finally, use 2t and the horizontal component of velocity to find the horizontal distance travelled.

Hope this helps=)

Answer 3

For the first problem, let's first look at the height as a function of time.
the speed in the upward direction is v0 - 9.8t
where v0 = 25*sin(35 degrees) and t is time in seconds.
The speed is zero when 25sin(35 degrees) = 9.8t, or
t = 25sin(35 degrees)/9.8
This is the time when the water is at its highest point.
Now we want to know where the water is horizontally, and that is easy. The horizontal distance is 25*cos(25 degrees)*t,
or 25*cos(25 degrees)*25*sin(35 degrees)/9.8
This is how far from the building you should locate the nozzle of the hose!

For the second problem, the first thing you want to do is figure out how long it takes the bullet to "fall" .025 meters. Then, multpliy this time by the horizontal speed of the bullet to get the horizontal distance the bullet has travelled. Use d = 1/2at^2 and solve for t: .025 = 1/2 * 9.8 * t^2
.025*2/9.8 = t^2
t = sqrt(.05/9.8)
Then the distance the bullet has traveled in this amount of time is
675*sqrt(.05/9.8)

The third problem is a little harder because you need to solve both equations at once, so let's write them down.
8.7 = horizontal speed * time in the air
= cos(23 degrees)*initial speed * t
and
t = sin(23 degrees)*initial speed*2./9.8
So now you have two equations, two unknowns and it's just algebra!

Answer 4

Yo, L -
just draw a picture for them. here i'll do #1 for you:

1. Let's say the stream of water has a velocity V and is fired at an angle of theta. The vertical component of the velocity is Vsin(theta) and the horizontal component is Vcos(theta). The vertical component of the velocity will decrease as the water moves to it's maximum height because of gravity. The horizontal component of the velocity will remain constant. So the question is asking, how far will the water have moved in the horizontal direction when it has reached the maximum height. Got it?

So it becomes pretty straightforward. The maximum height of the water occurs at time T where Vsin(theta) - gT = 0. Thus T = Vsin(theta)/g. Thus the horizontal movement is Vcos(theta)T = Vcos(theta)Vsin(theta)/g.

Plugging in the values we get:

25m/s * cos(35) * 25m/s * sin(35) / 9.8m/s^2 = 30.0 m

---

since no one is doing the other two:

#2: This one is pretty straightforward as well. The vertical component of velocity is going to increase as the bullet travels because of gravity. The distance is: d = 1/2 gT^2 where T is the time that passed. Thus, T = (2d/g)^.5. Knowing that the horizontal component of velocty doesnt change with time, we get the horizontal distance to be: 670m/s T. Plugging in we get:

670m/s * (2 * 0.025m / 9.8)^.5 = 47.9 m

#3. This one is very similar to #1. Using the same letter convention as number one we note that the total horizontal distance the dude jumps is 2TVcos(theta). NOTE that it's 2T not T because T is the time to get to the maximum height. The same amount of time is required for him to get back to the ground. As above, T = Vsin(theta)/g. So we have:

2Vsin(theta)Vcos(theta)/g = D.

Solving for V:
V^2 = Dg/(2sin(theta)cos(theta))

plugging in we have:

so V^2 = 8.7m * 9.8m/s^2 / (2 * sin(23) * cos(23))

= 118.5 m^2/s^2

so V = 10.9 m/s

Answer 5

AP Physics=cake...i got a 5 last year boooooy!