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