What is the initial velocity of the ball thrown by Nicole?

  • Thread starter Thread starter ae4jm
  • Start date Start date
  • Tags Tags
    Ball
Click For Summary
SUMMARY

The initial velocity of the ball thrown by Nicole is calculated to be 14 m/s. The problem involves a ball thrown straight up, passing a point 5.0 m above the release point with a downward speed of 10 m/s. The relevant equations used include Vf = Vi + at, y = Vi(t) + 0.5a(t^2), and Vf^2 = Vi^2 + 2(a)(y). The solution emphasizes the importance of correctly applying the signs in the equations of motion.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of acceleration due to gravity (-9.8 m/s²)
  • Ability to manipulate algebraic equations
  • Familiarity with concepts of projectile motion
NEXT STEPS
  • Study the derivation and application of kinematic equations
  • Learn about projectile motion and its characteristics
  • Explore the concept of energy conservation in motion
  • Practice solving similar problems involving vertical motion
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for practical examples in teaching these concepts.

ae4jm
Messages
79
Reaction score
0
[SOLVED] How fast was the ball thrown?

Homework Statement



Nicole throws a ball straight up. Chad watches the ball from a window 5.0m above the point where Nicole released it. The ball passes Chad on the way up, and it has a speed of 10 m/s as it passes him on the way back down. How fast did Nicole throw the ball?

Homework Equations


Vf=Vi + at
y=Vi(t)+.5a(t)(^2)
Vf^2=Vi^2+2(a)(y)


The Attempt at a Solution


I've tried starting with the final V =0 and I've tried to solve the downward motion of the ball first, to attempt to get maybe the y value, or height of the ball at it's max. Please, where should I start? What am I missing?
 
Physics news on Phys.org
A hint, since no energy is lost - the speed going back down at a point is the same as the speed going up at that same height.
You just need v^2 = u^2 + 2as be careful of the signs!
 
Thanks mgb_phys! I finally got the correct answer of initial velocity=14 m/s.

for v^2 = u^2 + 2as; I let the -10 m/s of the downward velocity, u, = + 10 m/s at the same height of 5m on the upward velocity. I let v = unknown variable and the acceleration was equal to -9.8 m/s^2, and s=5m for the height. Thanks for the explanation!
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
20
Views
2K
Replies
5
Views
8K
Replies
12
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
34
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K