A ball is thrown from the roof of a building

In summary, the ball is 980 meters away from the ground after it is thrown and the velocity is unknown.
  • #1
plissken170
5
0
It is s(t)=245+24.5t-4.9t(sqrd) meters from the ground (t) after it is thrown. s(t) = displacement function that describes the position of the ball from the ground at time t.

Determine height of building, time it takes the ball to reach the ground and the velocity when it hits the ground.I've been messing around with first, second derivatives and breaking down the original equation into t=10 and t=-5 but I don't know where to go from here.

Help apperciated.

i should mention i took physics 2 years ago and don't remember much of it
 
Last edited:
Physics news on Phys.org
  • #2
Well since the idea of negative time here is worthless, how could you use that t=10?
 
  • #3
Pengwuino said:
Well since the idea of negative time here is worthless, how could you use that t=10?

Well I plug it in the equation and get 980?
 
  • #4
What equation did you put t=10 into?
 
  • #5
s(t)=245+24.5t-4.9t(sqrd)
 
  • #6
Recalculate that. What do you think the position will be after 10 seconds?

You wouldn't think that after 10 seconds, the object is 980 meters off the ground...
 
  • #7
Of course I woudn't that's why it doesn't make any sense to me.
I think I am fundementally not getting some kind of simple concept here.
1st derivative when 10 is plugged in i get 1.47
 
  • #8
Well you were doing the calculation wrong.

245 + 24.5(10)-(4.9)(10)(10)=0 which is the position at t=10.

What equation do you use for finding the velocity in the y direction?
 
  • #9
v(t)=s dirvative(t)

I don't know. I'm not taking physics. I've covered my calculus books chapter and coudnt find anything.

I'm going to rest on it, thanks for you help though.
 
  • #10
Where is the ball at t=0?

At what time does s(t) = 0 and where is the ball then?

How do you find the function which discribes the velocity as a function of time?
 

1. What is the initial velocity of the ball?

The initial velocity of the ball is the speed at which the ball is thrown from the roof of the building. This velocity can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is the time.

2. How high does the ball go?

The maximum height the ball reaches can be calculated using the formula h = u^2/2g, where h is the maximum height, u is the initial velocity, and g is the acceleration due to gravity. However, this assumes no air resistance and a perfectly spherical ball. In reality, the height may vary slightly.

3. What is the acceleration of the ball?

The acceleration of the ball is due to gravity and is a constant value of 9.8 m/s^2. This means that the velocity of the ball changes by 9.8 m/s every second it is in the air.

4. How long does it take for the ball to hit the ground?

The time it takes for the ball to hit the ground can be calculated using the formula t = √(2h/g), where t is the time, h is the maximum height, and g is the acceleration due to gravity. This assumes no air resistance and a perfectly spherical ball.

5. What factors affect the trajectory of the ball?

The factors that affect the trajectory of the ball include the initial velocity, the angle at which the ball is thrown, air resistance, and external forces such as wind. The shape and weight of the ball may also play a role.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
34
Views
697
  • Introductory Physics Homework Help
Replies
31
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top