Projectile and acceleration of gravity

Click For Summary
SUMMARY

The discussion focuses on calculating the maximum height of a baseball hit in a pop-up scenario, where the ball travels 86 meters and remains airborne for 6.3 seconds under the acceleration of gravity at 9.8 m/s². The relevant equation used is y = vi(sin θ)t - 1/2gt², where θ is 90 degrees for a vertical launch. Participants clarify that the negative sign in the equation for gravity does not need to be repeated, and suggest breaking the problem into two parts: the ascent to the highest point and the descent back to the ground.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic knowledge of trigonometry, specifically sine functions
  • Concept of gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Calculate the time taken to reach the highest point using kinematic equations
  • Explore the derivation of projectile motion equations
  • Learn about the impact of initial velocity on projectile height
  • Investigate real-world applications of projectile motion in sports
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone interested in the mathematical modeling of sports dynamics.

armoredfury16
Messages
1
Reaction score
0

Homework Statement


During a baseball game, a batter hits a pop-
up to a fielder 86 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 6.3 s, how
high does it rise? Answer in units of m.

Homework Equations


y=vi(sin theta)t-1/2g(t)^2

The Attempt at a Solution



I'm not even sure I went about this right...I was only able to fill in that equation this much:

y=vi (sin90)6.3-1/2(-9.8?)(6.3)^2

As you can see, I'm pretty clueless as to how to find the height of the ball at it's highest point...
 
Physics news on Phys.org
Hi armoredfury16,

armoredfury16 said:

Homework Statement


During a baseball game, a batter hits a pop-
up to a fielder 86 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 6.3 s, how
high does it rise? Answer in units of m.

Homework Equations


y=vi(sin theta)t-1/2g(t)^2

The Attempt at a Solution



I'm not even sure I went about this right...I was only able to fill in that equation this much:

y=vi (sin90)6.3-1/2(-9.8?)(6.3)^2

Since the minus sign has been pulled out in front of the 1/2, you don't need it again with the 9.8.

If you want to use that equation, you might try writing it twice--once for the ball reaching the ground, and one for the ball reaching the highest point. What do you get? (In particular, how long does it take for the ball to reach the highest point?)
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
8
Views
3K
Replies
18
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
13
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K