Solve for Angle in 2D Projectile Motion Problem | Neglecting Air Resistance

[Georgeme_01]

Homework Statement

There are two cliffs separated by 40 meters (horizontal displacement). An arrow is shot from a bow at an angle and a velocity of 90 m/s. The arrow takes 0.75 seconds to arrive at the other cliff. What is the angle at which the arrow was released? Neglect air resistance.

Homework Equations

d_x = v_ix*t + 0.5*a_x*t²
Cos Θ = v_ix / v

The Attempt at a Solution

I used the equations and came up with 53.7°.
However, my question is why can I not solve for the initial velocity in the y direction using d_y = v_iy + 0.5*a_y*t² (where 2 is half of the original time) and solve for the displacement the arrow traveled in the y, then use v_fy² = v_iy² + 2a_yd
Using those formulas and using SinΘ, I calculate 2.34°.

They both can't be correct. Why is the second one incorrect?
And let me add that I'm the teacher that created this question, so it is possible that my question sucks. As I'm grading my student's papers, I'm wondering about this alternate calculation and I can't seem to find the flaw in the work. But it's not what I was expecting.

 

[paulfr]

There is no acceleration in the x direction.
So sx(t) = s0x + v0x t
40 = 0 + 90 cos [theta] (0.75)
cos [theta] = 0.593
theta = ArcCos 0.593 = 53.7 deg is correct

You would need the y direction information ONLY if you were not given
the time of travel. But it was given so you are saved that bit of computation.

If you were given the angle but not the time of travel, then you could calculate
the time for a round trip up and down using the y direction equation.

If you are not given the angle or the time of travel then you would need
both x and y direction equations which have angle and time as unknowns.
2 Eqs + 2 Unknowns ==> a solution

To answer your questions ...
1/ You already know the initial y velocity ... it is 90 sin [theta]
Acceleration is irrelevant for initial velocities
2/ Why solve for displacement ? It is 40m

 

[gneill]

Hi Georgeme_01, Welcome to Physics Forums.

However, my question is why can I not solve for the initial velocity in the y direction using d_y = v_iy + 0.5*a_y*t² (where 2 is half of the original time) and solve for the displacement the arrow traveled in the y, then use v_fy² = v_iy² + 2a_yd
Using those formulas and using SinΘ, I calculate 2.34°.

They both can't be correct. Why is the second one incorrect?

For your second method, what value did you use for V_iy? Seems to me that you have two unknowns to deal with, the maximum height of the trajectory and the initial vertical component of the velocity.

By the way, you're missing a "t" in your equation (shown in red):

d_y = v_iyt + 0.5*a_y*t²

 

[Georgeme_01]

I considered V_iy to be at the peak of the trajectory, considering the second half of the path. Knowing that my V_fy at the end of the path would be equal to the initial velocity at the beginning of the path.

 

[Georgeme_01]

You would need the y direction information ONLY if you were not given
the time of travel. But it was given so you are saved that bit of computation.

If you were given the angle but not the time of travel, then you could calculate
the time for a round trip up and down using the y direction equation.

If you are not given the angle or the time of travel then you would need
both x and y direction equations which have angle and time as unknowns.
2 Eqs + 2 Unknowns ==> a solution

To answer your questions ...
1/ You already know the initial y velocity ... it is 90 sin [theta]
Acceleration is irrelevant for initial velocities
2/ Why solve for displacement ? It is 40m

I'm still not sure I understand why it's wrong? I know the right way to do it, but why is solving using y-direction components, incorrect?

 

[gneill]

Actually, your first method won't give you the correct answer if you are expecting the arrow to travel only 40 m horizontally in total. It may reach the horizontal position of 40 m after 0.75 seconds, but it will still be high in the air at that point. Confirm using the range equation.

Your second method, assuming an "air time" of 0.75 seconds fixes the initial y-velocity, which in turn fixes the launch angle given the initial speed. That in turn gives you the x-velocity. Again the range will be larger than 40 m. Again, confirm with the range equation.

So it would appear that you can't satisfy all three of your given conditions at the same time (initial speed, horizontal range, and time of flight).

 

[PeroK]

I'm still not sure I understand why it's wrong? I know the right way to do it, but why is solving using y-direction components, incorrect?

You could get the $y$ component from $v_{iy} = \frac{gt}{2}$. But, that assumes the arrow arrives at the same height as it was fired.

 

[Georgeme_01]

Actually, your first method won't give you the correct answer if you are expecting the arrow to travel only 40 m horizontally in total. It may reach the horizontal position of 40 m after 0.75 seconds, but it will still be high in the air at that point. Confirm using the range equation.

At 0.75s it hits the other cliff (there is a person there that it hits). So yes, I know it's still high in the air, it's at the top of the other cliff.

Your second method, assuming an "air time" of 0.75 seconds fixes the initial y-velocity, which in turn fixes the launch angle given the initial speed. That in turn gives you the x-velocity. Again the range will be larger than 40 m. Again, confirm with the range equation.

I'm not sure I understand what you mean by "assuming an air time of 0.75s fixes the initial y-velocity, which in turn fixes the launch angle."

 

[Georgeme_01]

You could get the $y$ component from $v_{iy} = \frac{gt}{2}$. But, that assumes the arrow arrives at the same height as it was fired.

This gives the incorrect initial velocity in the y for my angle that I determined using x-components.

Why is this the incorrect initial velocity in the y? That is my question.

 

[PeroK]

This gives the incorrect initial velocity in the y for my angle that I determined using x-components.

Why is this the incorrect initial velocity in the y? That is my question.

Because if you want the arrow to go 40m in 0.75s and arrive at the height that it was fired, you need to fire it slower than 90m/s and probably at your angle of about 2.34°.

PS If you fire something at 90m/s at a low angle it will go approx 70m horizontally in 0.75s.

 

[Georgeme_01]

Because if you want the arrow to go 40m in 0.75s and arrive at the height that it was fired, you need to fire it slower than 90m/s and probably at your angle of about 2.34°.

PS If you fire something at 90m/s at a low angle it will go approx 70m horizontally in 0.75s.

But it can't travel 70m horizontally because at the same vertical height on the other side is a cliff with a person on it that the arrow is hitting.
Are you telling me the angle is actually 2.34° and so my work finding my initial horizontal velocity is incorrect??I'm not sure this forum is helping me solve my problem of explaining to my students why they got it wrong... or why I'm wrong in my thinking!

 

[gneill]

If it's fired at a low angle it'll arrive at the other side at the same height it was launched but the travel time will be much less than 0.75 seconds.

If it's fired at the higher angle it'll arrive at the other side in the correct time, but it'll much higher than the initial launch height when it reaches the desired horizontal distance.

In your first method you assumed that the distance covered was 40 meters in 0.75 seconds. That set the horizontal velocity but disregarded the y-velocity that the arrow is forced to take in order to apportion the fixed launch speed between the components. Fixing the horizontal speed does not fix the horizontal distance! The arrow will fly until it hits the ground, not until it reaches a set horizontal position. That horizontal position was only used to establish the horizontal velocity, not the height of the arrow at any particular time.

In your second method you assumed that the arrow flight time was 0.75 seconds, so reached the apex of its trajectory in half that time. Note that the horizontal range is ignored again. That gave you a maximum height of about 0.69 meters and an initial y-velocity of only about 3.7 m/s. The rest of the initial speed goes into the horizontal component and is about 89.9 m/s. The arrow is in the air for the right time but will travel nearly 70 meters horizontally. So again the range is wrong.

So really, both solutions are incorrect. There is no solution that can accommodate all three given conditions at the same time. Its like asking someone to find an angle in a triangle with side lengths 1, 2, and 12.

 

[Georgeme_01]

If it's fired at a low angle it'll arrive at the other side at the same height it was launched but the travel time will be much less than 0.75 seconds.

If it's fired at the higher angle it'll arrive at the other side in the correct time, but it'll much higher than the initial launch height when it reaches the desired horizontal distance.

In your first method you assumed that the distance covered was 40 meters in 0.75 seconds. That set the horizontal velocity but disregarded the y-velocity that the arrow is forced to take in order to apportion the fixed launch speed between the components. Fixing the horizontal speed does not fix the horizontal distance! The arrow will fly until it hits the ground, not until it reaches a set horizontal position. That horizontal position was only used to establish the horizontal velocity, not the height of the arrow at any particular time.

In your second method you assumed that the arrow flight time was 0.75 seconds, so reached the apex of its trajectory in half that time. Note that the horizontal range is ignored again. That gave you a maximum height of about 0.69 meters and an initial y-velocity of only about 3.7 m/s. The rest of the initial speed goes into the horizontal component and is about 89.9 m/s. The arrow is in the air for the right time but will travel nearly 70 meters horizontally. So again the range is wrong.

So really, both solutions are incorrect. There is no solution that can accommodate all three given conditions at the same time. Its like asking someone to find an angle in a triangle with side lengths 1, 2, and 12.

Thank you - this explanation really helped. I'll be changing this question to account for the actual horizontal displacement for the time that is given. In the meantime, I'll have to take both answers as long as the work for what was given is correct.

My students thank you :) This is what I get for trying to create my own unique questions.

 

[gneill]

I'm glad we could help. Cheers.