Question

A projectile is fired at v0 = 342.0 m/s at an angle of θ = 71.1o with respect to the horizontal. Assume that air friction will shorten the range by 32.1%. How far will the projectile travel in the horizontal direction, R?

Answer 1

@whpalmer4

Answer 2

@BAdhi

Answer 3

Calculate the horizontal and vertical components of velocity. Determine the time it will take to rise and fall due to gravity. Apply that time to horizontal speed then reduce it by the air friction.

Answer 4

What formula do I need to determine that?

Answer 5

\[Cos \theta=v_x/v_0\]\[\sin \theta=v_y/v_0\]

Answer 6

What would be the vx or vy?

Answer 7

they would be the horizontal (\(v_x\)) and vertical (\(v_y\)) components of the initial velocity vector \(\vec{v_0}=\langle v_x, v_y\rangle \)

Answer 8

So
x=342.0 m/s cos(71) = 111.3443088
y=342.0 m/s sin(71) = 323.3673529

Answer 9

@whpalmer4

Answer 10

@CShrix

Answer 11

Isn't the angle 71.1 degrees, not 71? Your x and y components will be off a little bit if that is true.

Answer 12

@gollywolly gave the right method, in my opinion. Find your \(v_y\) component first, so that you can write and solve the equation for the height of the projectile in order to find the time of flight. Then find the horizontal distance by multiplying the \(v_x\) component by time of flight, and reducing by the percentage mentioned for air resistance.

Answer 13

x=110.78 m/s
y=323.56 m/s

Answer 14

@whpalmer4 I got 75.21962m

Answer 15

@CShrix

Answer 16

hmm, you're going to shoot something straight up in the air at about 700 miles per hour, how quickly do you think that will come back to the ground?

Answer 17

I got 71.93s for the time.

Answer 18

Scratch that. Was looking at the wrong thing. I got 65.97s

Answer 19

if it is in the air for several dozen seconds, and it has a horizontal velocity of 110 m/s, how does it manage to only go 75 meters?!?

Answer 20

I get 66.0327 seconds for time of flight, using g = 9.8 m/s^2

Answer 21

Shouldn't it be -9.81m/s^2?

Answer 22

the magnitude is 9.8...the time changes to 65.9653 or 65.97 if I use 9.81, that's still not explaining your horizontal error :-)

If I didn't have the right sign on the first term, the two solutions for t would be 0 and -65.97 seconds...

Answer 23

The way I did it was
delta y =1/2(-9.81)detla t^2 +323.56 delta t
0=-4.905 delta t^2 +323.56 t
divide t from t^2 and t
0=-4.905t +323.56
-323.56=-4.905t
t=65.97s

Answer 24

not sure about your variable names, but the result is fine. so the projectile is up in the air for about 66 seconds, and traveling horizontally for each of the 66 seconds at 110.78 m/s. How did you get just 75.21962m for the distance traveled (prior to accounting for air resistance)?

Answer 25

0=1/2(0)(65.97)^2+110.78

Answer 26

0+110.78(.321)

Answer 27

110.78-35.56038

Answer 28

=75.21962m

Answer 29

Just follow the procedures that @whpalmer4 and @gollywolly gave!