Projectile Motion Problem (Physics 12) -- Apple core thrown from a tree

[fatcats]

Homework Statement

A child sitting in a tree throws his apple core from where he is perched (4.0m high) with a velocity of 5.0 m/s [35 degrees above horizontal], and it hits the ground right next to his friend.

A) How long is it before the apple core hits the ground?
B) How far from the base of the tree will the apple core land?
C) What is the velocity of the apple core on impact?

Homework Equations

d = v1 * t + 1/2 a (t)^2
a = v2-v1 / t

(I included a diagram in this post)

The Attempt at a Solution

I got these answers. I googled this problem and my answers all wrong but I don't understand how to find the right answer.

a) 0.65 seconds
I got this by solving for the component vy (I got 2.9 m/s) and then plugging it into my first relevant equation and solving for t using quadratic formula.

b) I got 2.7m for this. I got this by solving for vx (4.1m/s) and then plugging vx and time into the first relevant equation I posted. Maybe my method was right for this, I'm not sure, but I know that I got time wrong so the number didn't come out right...

c) I used the second relevant equation for this and got 9.3 m/s. Again I know that my time value was wrong, so I assume that is probably what skewed this instead of a wrong methodology.

I would really like to understand the material so clarification on what I did wrong (or didn't do at all), especially in part a) would be greatly appreciated.

Thanks

 

[stockzahn]

@ a): I obtained a different solution for the time. Which values do you use for the variables in your first relevant equation?

 

[fatcats]

Hi there,

Here is a breakdown of my process.

1) sin35 = vy/5
vy = 5sin35
vy = 3.9

Then for the first equatio:

4 = 2.9 (t) + 1/2(9.8)(t)^2
I simplified that to get:
0 = 4.9t^2 + 2.9t - 4

Then I finished with t = -1.2 and t = 0.65 seconds.

 

[stockzahn]

Hi there,

Here is a breakdown of my process.

1) sin35 = vy/5
vy = 5sin35
vy = 3.9

Then for the first equatio:

4 = 2.9 (t) + 1/2(9.8)(t)^2
I simplified that to get:
0 = 4.9t^2 + 2.9t - 4

Then I finished with t = -1.2 and t = 0.65 seconds.

What would be the time needed for the apple to reach the ground if you just would drop it?

 

[PeroK]

... or the time if the angle was 35° below the horizontal?

 

[fatcats]

I don't understand how to solve for the time needed for the apple to reach the ground just by dropping it... or what it would change.
Isn't it still the same distance traveled for dy (4 m)?

Alright so if the angle is 35 below horizontal. Does that mean an angle of 325?

5 * sin 325 = -2.87
So the same value but in the negative.

Therefore when I apply it in quadratic formula I get the same numbers but with reversed signs, so 1.2 and -0.65.

That was all I did wrong (excepting the wrong time value for b and c)?

Thank you for the help

 

[haruspex]

I don't understand how to solve for the time needed for the apple to reach the ground just by dropping it... or what it would change.

I believe stockzahn was encouraging you to compare your answer with the time taken if it were simply dropped. You would have found that time was greater than you had calculated for its being thrown at angle upwards, which would make no sense.

 

[fatcats]

d = v1 * t + 1/2 a (t)^2
4 = 0 * t + 1/2 (9.8) t ^2
4.8t^2 - 4 = 0
I did quadratic theory after using a = 4.8, b=0, and c=-4. I got +-0.89... so if I added an initial velocity wouldn't the correct time be faster than that? Or is it that because it is thrown at an angle upwards it has to reach a highest point and then fall so it takes more time to land?

I still don't understand how to get the correct answer, was changing the angle to 35 below horizontal (which I put in as sin325) the correct way to solve this problem?

 

[haruspex]

because it is thrown at an angle upwards it has to reach a highest point and then fall so it takes more time to land?

Yes, but more convincingly: it takes time to reach the highest point; having got there it has no vertical velocity so just drops, but it now has further to go than initially, so must take longer.

I still don't understand how to get the correct answer, was changing the angle to 35 below horizontal (which I put in as sin325) the correct way to solve this problem?

Not exactly You need to be consistent with signs.
First, decide whether up is positive or negative. If positive, the initial vertical speed is positive, 5 sin(35), but the acceleration and displacement are negative. If up is negative, acceleration and displacement are positive but the initial velocity is at angle -35 degrees to the horizontal.

 

[fatcats]

Okay, thank you for the clarification. I had designated that downwards was positive in my original answer. I understand it now.

 

[fatcats]

Hi, I have one last question regarding this.

So after finding the proper value for time I found the right answer for part b. dx=4.9

For c, I found the correct velocity, 9.8 m/s. Then I also found the angle to be 65 degrees or 245 degrees. I have to make a concluding statement and if it is not against this forum's rules, please let me know if my logic on this makes sense:

"Therefore the final velocity is 9.8 m/s 65 degrees below horizontal. This is because the 245 value falls below the 180 degree line."

I did this: tan angle = 8.9/4.1 and rearranged for the angle.

 

[haruspex]

65 degrees below horizontal. This is because the 245 value falls below the 180 degree line."

Not quite sure what you mean by "the 180 degree line". Yes, 245 exceeds 180, so would mean the core is now moving upwards, so you can rule that out.
Fwiw, the ambiguity arises because the magnitude is found through a square root, implicitly producing two solutions of opposite sign. The negative value would correspond to the 245 degree angle, which solution is completely equivalent to the positive root with the 65 degree angle.

 

[fatcats]

Er, I meant the horizontal by the 180 degree line.

I understand what I did now... I designated bottom and forwards as positive. Therefore my +65 angle is actually below horizontal ^^"

Thanks for getting back to me!

 

[No1_129848]

One big important thing, your formula for a is the (right) formula for an average acceleration.

Do you understand what is the (only) acceleration you should worry about in this specific problem?

Once you setup your correct equation of motion (and ofc you can derive them if you need eqs for velocity and whatnot) you should think about the t you want to obtain from them, and what you know it has to be true in that specific t.
In this way you should be able to solve one of your equation ot the t the apple lands on the ground, once you have t you can easily find both the vf and the xf by plugging it in your equations.
Also, do the right thing and start considering whatever you can as a vector, your equations are vectorial so you have to be careful with signs and possible angles.
Not doing so from the beginning will result in nightmares once you start with dynamic

 

[fatcats]

It's just the downwards acceleration of gravity, right? 9.8

It's just the equation provided to me for this lesson centered around projectile motion in grade 12 physics