Projectile Motion Problems: Solving for Time with Given Initial Speed and Height

  • Thread starter Sophialiu609
  • Start date
In summary, to solve for the time it takes for a projectile to hit the ground after being fired upward with an initial speed of 2940 m/s, use the equation h=rt-4.9t^2 and solve for t. Remember to convert the time to minutes if necessary. For the second problem, use the same equation and solve for t to find the time it takes for a ball to hit the ground after being thrown upward from the top of a 98m tower with an initial speed of 39.2 m/s. Make sure to convert the time to minutes as well.
  • #1
Sophialiu609
10
0
t=number of seconds
h=height
r=initial speed
How do you solve these two problems?
1. A projectile is fired upward with an initial speed of 2940 m/s. After how many minutes does it hit the ground?
2. A ball is thrown upward from the top of a 98m tower with an initial speed of 39.2 m/s. When does it hit the ground?
I don't understand how you can find the answers to these types of problems
 
Physics news on Phys.org
  • #2
Help PLEASE!
I got this far on the second one...
h=rt- 4.9t²
-98=39.2t-4.9t²
-20=8t-t²
0=-(t²-8t -20)
0=-(t-10)(t+2)
(t-10)=0 (t+2)=0
t=10 t=-2
 
Last edited:
  • #3
For the first:
If you don't need the height then why are you calculating it ?
Try Vf = Vi + at
Where t would be time of flight & Vf,Vi final and initial velocity respectively
 
Last edited:
  • #4
Well, I'm supposed to put the problem into that formula, that's what I'm confused about.
 
  • #5
Don't forget to convert time to minutes b/c the answer you will get is in seconds.
 
  • #6
For the first, if it's fired from the ground and lands on the ground, what is h (often times, more appropriately written as [tex]\Delta[/tex]h)?
 
  • #7
Sophialiu609 said:
Well, I'm supposed to put the problem into that formula, that's what I'm confused about.

For the first question: you can use the same equation as you did for problem #2. If the projectile returns to the ground, what is its displacement? What does that make h in your equation?
 
  • #8
Ok, so what I have so far is
0=2940t/60 -4.9t^2/60
0=-4.9t(t/60 -1)
0=-4.9t(t-1/60)
0=-4.9t(t-1)
t={0,1}
Is that right?
 
  • #9
0=2940t-4.9t²
60 60
0=2940t-4.9t²
0=600t-t²
0=-t(t-600)
-t=0 (t-600)=0
t=0 t=600

or is this right?
 
  • #10
Just a little tip, you might want to work through things in algebra before you number plug :)
 
  • #11
Would time 0 make sense as an answer?

And what units are your answers in?
 
  • #12
0 wouldn't make sense as an answer, my answer is in minutes~I think... but now I'm more confused...
 
  • #13
Is it 10 minutes...?
 
  • #14
Sophialiu609 said:
Is it 10 minutes...?
Stop guessing. What equation are you using?

And before plugging values into your equation, let's SOLVE for the variable we want.
 
  • #15
The one I posted as the title!
h=rt-4.9t^2
 
  • #16
Sophialiu609 said:
The one I posted as the title!
h=rt-4.9t^2
Well looks right to me. You have it in terms of minutes, so there you go.
 
  • #17
So was my answer correct? 10 minutes? Because 600/60=10
 
  • #18
oh ohkay, thank you i didn't see the second page
 

1. What is H=rt-4.9t^2?

H=rt-4.9t^2 is a mathematical equation that represents the height (H) of an object at a certain time (t) when thrown or dropped with an initial velocity (r) and under the force of gravity.

2. How is H=rt-4.9t^2 used in science?

This equation is commonly used in physics and engineering to calculate the height of objects in free fall or projectile motion. It can also be used to analyze the trajectory of objects and predict their landing position.

3. What is the significance of the "-4.9t^2" term in H=rt-4.9t^2?

The "-4.9t^2" term represents the acceleration due to gravity (g), which is approximately 9.8 meters per second squared (m/s^2). This value is negative because gravity acts in the downward direction, causing objects to accelerate towards the ground.

4. Can H=rt-4.9t^2 be used for objects thrown upwards?

Yes, this equation can be used for objects thrown upwards, as long as the initial velocity (r) is also positive. However, the values for time and height will be negative, as the object will eventually fall back down to the ground.

5. How does air resistance affect the accuracy of H=rt-4.9t^2?

If an object is experiencing air resistance, the value of r (initial velocity) will change over time, making the equation less accurate. This is because air resistance acts in the opposite direction of the object's motion, slowing it down. For more accurate results, a more complex equation that takes air resistance into account may be used.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
230
  • Introductory Physics Homework Help
Replies
3
Views
576
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
691
  • Introductory Physics Homework Help
Replies
11
Views
871
  • Introductory Physics Homework Help
Replies
4
Views
906
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
388
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top