Do not understand the question: Projectile motion (got correct answers)

[Apollinaria]

So I got the correct answers for this question but I don't understand how.
I don't understand the breakdown of the components.
I don't understand where what variables should be.

Homework Statement

A rock is projected at a cliff with initial speed of 42m/s, at 60deg above the horizontal. It strikes A at 5.5s.

Find the height of the cliff (I labeled it as dy but don't know if that's the same thing).
Find the speed before impact at A.
Find the max height reached (H).

The Attempt at a Solution

Attached.
Why is gravity negative?
Why do we find dy instead of H and how did this happen?
Is there no acceleration in the x direction at max height? So a=0? That's the only way it works out.
Why doesn't my first solution for c) work?

I don't understand the concept behind this problem at all.

 

[Bill Nye Tho]

Gravity is negative depending on your reference frame. If up is in the positive Y direction then down will be negative.

I'm not sure that I under your second question, so correct me if I don't explain it right but you find "dy" because you plug in 5.5s for your time. You are not given the time it takes for it to reach H but you do know that your velocity in the vertical component will be 0 at the very moment that the object reverses it's vertical direction.

One of the key points of projectile motion is that your horizontal velocity is constant therefore your acceleration is always = 0. (Ignoring drag/etc)

 

[Apollinaria]

Gravity is negative depending on your reference frame. If up is in the positive Y direction then down will be negative.

I'm not sure that I under your second question, so correct me if I don't explain it right but you find "dy" because you plug in 5.5s for your time. You are not given the time it takes for it to reach H but you do know that your velocity in the vertical component will be 0 at the very moment that the object reverses it's vertical direction.

One of the key points of projectile motion is that your horizontal velocity is constant therefore your acceleration is always = 0. (Ignoring drag/etc)

I had no idea that a=0 for projectile motion. Thanks for that bit of info.
So you're saying that if I had a time for H, I could plug that into find H?
I need to rephrase my own question..

Why is 42sin60 Viy? Is Viy same as H? Or same as Diy? I wasn't sure what I was finding there. What is Viy anyway?

 

[SteamKing]

Viy is the component of the initial velocity of the projectile in the y-direction.

 

[Bill Nye Tho]

I had no idea that a=0 for projectile motion. Thanks for that bit of info.
So you're saying that if I had a time for H, I could plug that into find H?
I need to rephrase my own question..

Why is 42sin60 Viy? Is Viy same as H? Or same as Diy? I wasn't sure what I was finding there. What is Viy anyway?

Your acceleration is equal to zero only in the horizontal component.

Your vertical component of acceleration will be gravity.

Exactly, but when you're asked to find the maximum height of a projectile (H), they generally want to see if you understand that your vertical component of velocity will be zero for that sudden moment when it arrives at H.

You are given that your velocity is 42 m/s @ 60 degrees above the horizontal. You need the x and y components of that velocity in order to calculate. You use trigonometry to find the components of your velocity. The sine of 60 degrees multiplied by 42 m/s will give you the Y component of your velocity and the cosine of 60 degrees multiplied by 42 m/s will give you the X component of your velocity.

H and ViY are not the same.

Remember: H is the maximum height that the projectile will reach before it starts being pulled back down and ViY is the Y component of your velocity.

 

[Apollinaria]

Your acceleration is equal to zero only in the horizontal component.

Your vertical component of acceleration will be gravity.

Exactly, but when you're asked to find the maximum height of a projectile (H), they generally want to see if you understand that your vertical component of velocity will be zero for that sudden moment when it arrives at H.

You are given that your velocity is 42 m/s @ 60 degrees above the horizontal. You need the x and y components of that velocity in order to calculate. You use trigonometry to find the components of your velocity. The sine of 60 degrees multiplied by 42 m/s will give you the Y component of your velocity and the cosine of 60 degrees multiplied by 42 m/s will give you the X component of your velocity.

H and ViY are not the same.

Remember: H is the maximum height that the projectile will reach before it starts being pulled back down and ViY is the Y component of your velocity.

So Vfy will be zero because the ball briefly stops before it comes down. And Viy is what we initially found for the y component (42sin60). I'm just confused as to why we're not using 42m/s as our Vi instead of the y component. Ughhh :grumpy: