Calculating How High a Softball Thrown at 17.5m/s Rises

  • Thread starter simpleee
  • Start date
In summary, when a softball is thrown at a speed of 17.5m/s, the height it will reach can be calculated using the formula h = (v^2)/(2g), where h is the height and v is the initial velocity of the softball. This formula takes into account the acceleration due to gravity (g) and can be used to determine the maximum height the softball will reach before falling back to the ground.
  • #1
simpleee
13
0
A softball is thrown straight up at 17.5m/s and caught at 3.60s later.How high does the ball rise?

What equation would I use for this?
I know that:
Vi = 17.5m/s
t = 3.6s
and gravity is -9.8
Can anyone help me get started on this problem?
 
Physics news on Phys.org
  • #2
[tex]s(t)=x_{i}+v_{i}t+\frac{1}{2}at^{2}[/tex]

Also, the total elapsed time is 3.6s. Remember that this is the time it takes for the ball to reach it's highest point and then return.
 
  • #3
what does xi equal?
 
  • #4
s(-9.8) = xi + 17.5 - 4.9t

[I think]
 
  • #5
[tex]x_{i}[/tex] just represents your initial position. In your case it is zero.

Use the equation I posted, this calculates displacement. You know that a = -9.8 and your initial velocity is 17.5, lastly you know your initial height is zero. Put that information in the general equation and then calculate the time where the ball would be highest.
 
  • #6
If the round trip takes 3.6 seconds, then at what time will the ball be at it's highest point?
 
  • #7
Here's a hint. When an object is thrown up, it spends half its time going up and the other half coming back down.
 
  • #8
I'm not getting it.
whats s stand for?
when I plug it in it is: s(-9.8) = 0 + 17.5 - 4.9(3.6)
 
  • #9
s(-9.8) = 0 + 17.5 - 4.9(3.6)
s(-9.8) = 0 + 17.5 - 17.64
s(-9.8) = 17.5 - 17.64
?
 
  • #10
It takes the same amount of time to go up as it does to come back down...
 
  • #11
Would it make more sense if I said [tex]
f(t)=v_{i}t+\frac{1}{2}at^{2}
[/tex]

f(t) is just your output; this equation represents the ball's displacement as a function of time. Also keep in mind that the acceleration is multiplied by [tex]t^{2}[/tex] and your initial velocity is multiplied by t.

Where are you getting s(-9.8) from?

[tex]
f(t)=17.5t-4.9t^{2}
[/tex]

Thats your equation. Now do you see why the time you need to substitute into that function is not 3.6?
 
  • #12
simpleee said:
s(-9.8) = 0 + 17.5 - 4.9(3.6)
s(-9.8) = 0 + 17.5 - 17.64
s(-9.8) = 17.5 - 17.64
?

No, you're trying too hard.
 
  • #13
Still confused?
 

1. How do you calculate the height a softball reaches when thrown at 17.5m/s?

To calculate the height a softball reaches when thrown at 17.5m/s, we can use the equation h = (v2sin2θ) / 2g, where h is the height, v is the initial velocity of the softball, θ is the angle of the throw, and g is the acceleration due to gravity (9.8m/s2).

2. What is the initial velocity of a softball thrown at 17.5m/s?

The initial velocity of a softball thrown at 17.5m/s is 17.5m/s. This is the speed at which the softball leaves the thrower's hand.

3. Can the height a softball reaches be affected by the angle of the throw?

Yes, the height a softball reaches can be affected by the angle of the throw. The higher the angle of the throw, the higher the softball will rise. This is because a higher angle results in a greater vertical component of the initial velocity.

4. How does the acceleration due to gravity affect the height a softball reaches?

The acceleration due to gravity affects the height a softball reaches by pulling the softball towards the ground. As the softball rises, it slows down due to the gravitational pull, and then starts to fall back towards the ground. The greater the acceleration due to gravity, the shorter the time the softball is in the air and thus the lower the height it reaches.

5. What is the maximum height a softball can reach when thrown at 17.5m/s?

The maximum height a softball can reach when thrown at 17.5m/s is dependent on the angle of the throw. However, assuming a 45-degree angle, the maximum height would be 12.65 meters. This is the highest point the softball reaches before falling back towards the ground.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
999
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Back
Top