Kinematics: Projectiles

[AnaB]

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I'm having trouble with a kinematics question -

If two projectiles are launched with the same speed but different angles of 30 and 60 degrees. How do their flight times and horizontal distances traveled compare?

Apparently the horizontal distances are the same, but their flight times differ.

The explanation they provide is convoluted and makes no sense, so if anyone can present this to me simply, it would be very appreciated.

 

[WhiteWashed]

I'm having trouble with a kinematics question -

If two projectiles are launched with the same speed but different angles of 30 and 60 degrees. How do their flight times and horizontal distances traveled compare?

Apparently the horizontal distances are the same, but their flight times differ.

The explanation they provide is convoluted and makes no sense, so if anyone can present this to me simply, it would be very appreciated.

while flight times vary, so do the y axis velocities. at 30 degrees, you will probably have a shorter flight time, but faster y axis velocity. at 60 degrees, you will have a longer flight time, but shorter y axis velocity.

Its like 3x5=15 and 5x3=15 just flipped flopped.

also i remember reading something about how complimentary angles are often end up with the same answers i.e. 30 & 60, 20 & 70 blah blah

hope that makes sense.

 

[AnaB]

while flight times vary, so do the y axis velocities. at 30 degrees, you will probably have a shorter flight time, but faster y axis velocity. at 60 degrees, you will have a longer flight time, but shorter y axis velocity.

Its like 3x5=15 and 5x3=15 just flipped flopped.

also i remember reading something about how complimentary angles are often end up with the same answers i.e. 30 & 60, 20 & 70 blah blah

hope that makes sense.

Hmm although your explanation makes sense intuitively, I think its contradicting what the solutions say - you are implying that the flight times also vary and according to the explanation thats not the case.

When I answered this question, I thought that the flight times would be the same, but the distances traveled would differ.

Here is the explanation they provide, maybe it will help...
R depends on sin 2. Since sin 2(30°) = sin 2(60°), the projectiles’ ranges will be the same; this eliminates choices B and C. The total flight time, T, is found by doubling the time t required for the projectile’s vertical velocity to drop to 0 (which occurs at the projectile’s maximum height). Since Dvy= –gt, we have t = v0y/g, so T = 2v0y /g = 2(v0 sin )/g. Since sin 30°sin 60°, the projectiles’ flight times will not be the same, so the answer is D.

(This is TPR Exam 6, Q33)

 

[WhiteWashed]

Indeed that is very convulted. however the answer does state that the flight time varies....Since sin 30°sin 60°, the projectiles’ flight times will not be the same, so the answer is D.

Hmm although your explanation makes sense intuitively, I think its contradicting what the solutions say - you are implying that the flight times also vary and according to the explanation thats not the case.

When I answered this question, I thought that the flight times would be the same, but the distances traveled would differ.

Here is the explanation they provide, maybe it will help...
R depends on sin 2. Since sin 2(30°) = sin 2(60°), the projectiles’ ranges will be the same; this eliminates choices B and C. The total flight time, T, is found by doubling the time t required for the projectile’s vertical velocity to drop to 0 (which occurs at the projectile’s maximum height). Since Dvy= –gt, we have t = v0y/g, so T = 2v0y /g = 2(v0 sin )/g. Since sin 30°sin 60°, the projectiles’ flight times will not be the same, so the answer is D.

(This is TPR Exam 6, Q33)

 

[tartrate]

at 30 degrees, you will probably have a shorter flight time, but faster y axis velocity. at 60 degrees, you will have a longer flight time, but shorter y axis velocity.

This is incorrect. At 60 degrees you will have a longer flight time AND larger y-velocity (think of it this way, since x and y velocity vectors are independent, why else would you have a longer flight time?). And of course, you will have maximum flight time and y-velocity at 90 degrees.

As to understanding the question, it works out mathematically if you work through it. Or you can just think of it as trading between time in air (related to y-velocity) for time traveling horizontally (related to x-velocity).

But seeing the math helps sometimes:

x=v*cosA*t_total

g=v*sinA/t, time is for either for launch to peak height or vice versa

t_total=2*v*sinA/g (time above is multiplied by 2 for total time in air)

x=2*v^2*sinA*cosA/g (just substitution)
=v^2*sin2A/g (a bit of trigonometric manipulation, but not necessary)

Both sin2*30=sin60 and sin2*60=sin120 are equal, so x is equal in both cases.

Also, if the last step is elusive, you can also compare sin30cos30 to sin60cos60. You will see they come out to be the same value.

And for my very own clever little shortcut I just came up with: in the end, all you need to do is compare sin2A for two different angles to see what travels further.

And now that I think of it even more, you can actually use the last equation for x to find the maximal range using calculus!

 

[WhiteWashed]

Sorry, i typed Y axis, when i meant X axis.

It should read,

while flight times vary, so do the y axis velocities. at 30 degrees, you will probably have a shorter flight time, but faster X axis velocity. at 60 degrees, you will have a longer flight time, but slower X axis velocity.

Its like 3x5=15 and 5x3=15 just flipped flopped.

also i remember reading something about how complimentary angles are often end up with the same answers i.e. 30 & 60, 20 & 70 blah blah

hope that makes sense.

This is incorrect. At 60 degrees you will have a longer flight time AND larger y-velocity (think of it this way, since x and y velocity vectors are independent, why else would you have a longer flight time?). And of course, you will have maximum flight time and y-velocity at 90 degrees.

As to understanding the question, it works out mathematically if you work through it. Or you can just think of it as trading between time in air (related to y-velocity) for time traveling horizontally (related to x-velocity).

But seeing the math helps sometimes:

x=v*cosA*t_total

g=v*sinA/t, time is for either for launch to peak height or vice versa

t_total=2*v*sinA/g (time above is multiplied by 2 for total time in air)

x=2*v^2*sinA*cosA/g (just substitution)
=v^2*sin2A/g (a bit of trigonometric manipulation, but not necessary)

Both sin2*30=sin60 and sin2*60=sin120 are equal, so x is equal in both cases.

Also, if the last step is elusive, you can also compare sin30cos30 to sin60cos60. You will see they come out to be the same value.

And for my very own clever little shortcut I just came up with: in the end, all you need to do is compare sin2A for two different angles to see what travels further.

And now that I think of it even more, you can actually use the last equation for x to find the maximal range using calculus!

 

[TwoPaddles]

I'm having trouble with a kinematics question -

If two projectiles are launched with the same speed but different angles of 30 and 60 degrees. How do their flight times and horizontal distances traveled compare?

Apparently the horizontal distances are the same, but their flight times differ.

The explanation they provide is convoluted and makes no sense, so if anyone can present this to me simply, it would be very appreciated.

Think of it this way..time of flight refers to the y-direction.
You are given 30 degrees and 60 degrees. So we would need to look at sin30 and sin60 (since sin is for the y-direction in a projectile).

Let's try unusual cases:
What if you had 0 degrees? Its flight time would be zero because it will not be on the air.
What if you had 90 degrees? It will have the largest flight time because you are shooting the projectile straight up.
sin(0) = 0
sin(90) = 1

Therefore, the closer the angle to the "straight up" value will stay in flight longer.
sin(30) = 0.5
sin(60) = 0.87

So the angle with 60 degrees will stay in flight longer.


As far as the horizontal distance (range),
Range = v*t, where v is the velocity in the horizontal direction.
v in horizontal is:
cos(30) = 0.87
cos(60) = 0.5

so the angle with 30 degrees will have a longer range because its v (in x direction) is larger.