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

AI Thread Summary
The discussion focuses on calculating the vertical distance (deltaH) of a projectile fired from a toy cannon, specifically at time t = 0.615 seconds. The initial velocity and angle are provided, and initial attempts to calculate deltaH using vertical and horizontal components of motion were rejected. The key insight is that the problem requires finding the difference in vertical positions between a straight-line trajectory and a parabolic path influenced by gravity. The correct formula for deltaH is derived as (1/2)gt^2, which represents the vertical distance fallen due to gravity over the specified time. This highlights the importance of understanding the effects of gravity on projectile motion in two dimensions.
Agent M27
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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


yf=yi+ vyit+1/2ayt2

xf=xi+ vxit+1/2axt2

vyi=visin\vartheta

vxi=vicos\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:

vyi=8.9m/s(sin56)=7.378434 m/s
vxi=8.9m/s(cos56)=4.976816 m/s

\Deltah=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 vias the hypotenuse and x was found using the following:

xf=vxit
=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


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

Agent M27 said:
\Deltah=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?
 
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
 
Agent M27 said:
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:

ystraight = vy0t

and

yparabolic =vy0t - (1/2)gt2

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

ystraight - yparabolic = vy0t - [vy0t - (1/2)gt2 ]

= (1/2)gt2
 

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