How Do You Calculate Projectile Motion in Physics?

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To calculate projectile motion, the discussion centers on a stone projected at a cliff with an initial speed of 40.0 m/s at a 56.0° angle, striking the cliff after 5.10 seconds. The participants explore how to find the maximum height, emphasizing that at this point, the vertical velocity is zero. The vertical motion equation, Vy = Vi*sin(angle) - gt, is highlighted as essential for determining the stone's behavior. It is clarified that the time to reach maximum height is not simply half the total time due to differing initial and final heights. The conversation concludes with understanding how to apply the equations correctly to solve for the required variables.
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n Fig. 4-33, a stone is projected at a cliff of height h with an initial speed of 40.0 m/s directed at an angle 0 = 56.0° above the horizontal. The stone strikes at A, 5.10 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.

Part c is basically asking for the trajectory, right?
 
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frankfjf said:
Part c is basically asking for the trajectory, right?
In a sense, yes, but only one aspect of it. How would you go about finding the maximum height?
 
To be honest I'm not sure.
 
Here's a hint: What's the vertical component of the velocity when the stone reaches its highest point?
 
That's just it, I have no idea what to plug in for determining the velocity at the highest point.

I think that the formula involved would be the vertical motion equation for velocity, Vy = Vi*sin(angle) - gt.

Well actually, if I took half of the time, wouldn't it be at its highest point then?
 
frankfjf said:
That's just it, I have no idea what to plug in for determining the velocity at the highest point.
Think about it. Imagine throwing a ball straight up in the air. Initially it moves up (velocity = +); after reaching the highest point, it starts moving down (velocity = -). So what can you say about the speed at the highest point?

I think that the formula involved would be the vertical motion equation for velocity, Vy = Vi*sin(angle) - gt.
Good.

Well actually, if I took half of the time, wouldn't it be at its highest point then?
Only if the initial and final positions were at the same height, which is not true in this problem.
 
Doc Al said:
Think about it. Imagine throwing a ball straight up in the air. Initially it moves up (velocity = +); after reaching the highest point, it starts moving down (velocity = -). So what can you say about the speed at the highest point?

Well, at the highest point the speed would be zero, since it's going to start falling immediately after hitting the highest point.
 
Exactly right. Now make use of that fact.
 
But now I'm confused as to just where to plugin the zero.

If the formula Vy = Vi*sin(angle) - gt is used, does the zero replace Vy or Vi?
 
  • #10
What do you think? (Does the initial velocity change?)
 
  • #11
No, but if I put in zero for Vy, I'm left without any variables.
 
  • #12
You can solve for the time; that's what you need.
 
  • #13
Ahhh I got it now, thanks!
 
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