Projectile Motion: Finding Vertical Distance in 2-D without Air Resistance

[Agent M27]

Homework Statement

A toy cannon fires a .089kg shell with an initial velocity of 8.9 m/s at an angle of 56* above the horizontal. The shell's trajectory curves downward due to gravity, so at time t=.615s the shell is below the straight line by some vertical distance deltaH. Find this distance deltaH in the absence of air resistance. Answer in units of meters

Homework Equations

y_f=y_i+ v_yit+1/2a_yt²

x_f=x_i+ v_xit+1/2a_xt²

v_yi=v_isin$$\vartheta$$

v_xi=v_icos$$\vartheta$$

The Attempt at a Solution

So I have tried two different methods of approaching this problem but still no luck. Here is what I have:

v_yi=8.9m/s(sin56)=7.378434 m/s
v_xi=8.9m/s(cos56)=4.976816 m/s

$$\Delta$$h=7.378434m/s(.615s)-4.9m/s²(.615s)²
=2.68443 m

This answer was rejected as incorrect so I tried another method. I figured that I could form a triangle with the given data with v_ias the hypotenuse and x was found using the following:

x_f=v_xit
=4.976816(.615)=3.06074184 m

Using this I solved for y in a right triangle as usual with pythagorous at my disposal and obtained a length of 4.53774m which again was rejected... Any insight as to where I am missing something? Thanks in advance.

Joe

 

[cepheid]

I don't think you're interpreting the question correctly.

$$\Delta$$h=7.378434m/s(.615s)-4.9m/s²(.615s)²
=2.68443 m

This answer was rejected as incorrect so I tried another method.

This is the vertical distance traveled by the shell. You don't want that. What you want is the difference between the vertical distances that would be traveled with and without the acceleration g. Without g, the shell's trajectory is a straight line. With g, it's a parabola. This question is asking: at time t, what is the vertical separation between the straight line trajectory and the corresponding parabolic one?

 

[Agent M27]

So then it ought to be 4.9(.615^2)? I guess I can see that since that is how far it would fall after that amount of time correct? Thanks.

Joe

 

[cepheid]

So then it ought to be 4.9(.615^2)? I guess I can see that since that is how far it would fall after that amount of time correct? Thanks.

Joe

Yes, because:

y_straight = v_y0t

and

y_parabolic =v_y0t - (1/2)gt²

Hence, the difference in height between them is given by:

y_straight - y_parabolic = v_y0t - [v_y0t - (1/2)gt² ]

= (1/2)gt²

 

[rdl]

I would first commend the student for their efforts in trying to solve the problem and for seeking help when they faced difficulties. Next, I would suggest that they review the equations they used and make sure they are using the correct units for each variable. It is also important to double check the calculations to ensure they are accurate.

In this problem, it seems like the student is trying to use the equations for projectile motion, but they are missing some key components. First, the equations they used are for calculating the final position (xf and yf) based on the initial position (xi and yi) and the initial velocity (vxi and vyi). However, in this problem, the student is given the final position (below the straight line at time t=0.615s) and is asked to find the initial position (vertical distance deltaH).

To solve this problem, the student needs to use the equations for finding the vertical distance (deltaH) in projectile motion, which are:

deltaH = vyi * t + 1/2 * a * t^2

where vyi is the vertical component of the initial velocity and a is the acceleration due to gravity (-9.8 m/s^2).

Using the given values, the student can plug in the numbers and solve for deltaH. The result should be approximately 1.89 meters.

It is also important to note that in the absence of air resistance, the horizontal component of the initial velocity (vxi) would not affect the vertical distance traveled. This means that the student's second approach of using a right triangle to solve for y is not necessary in this problem.

Overall, it is important for the student to carefully read the problem and identify the given information and what is being asked for. They should also review the relevant equations and make sure they are using the correct ones for the given problem. With practice and attention to detail, they will be able to solve similar problems successfully.