Projectile Motion at an Angle: Solving for Velocity, Time, and Height

Click For Summary

Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a ball thrown at an angle. The original poster presents a scenario where a ball is thrown with an initial velocity of 15 m/s at a 60-degree angle, and seeks to determine the velocity at maximum height, the time to reach that height, and the maximum height itself.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the horizontal and vertical components of the initial velocity and questions the implications of having no vertical velocity at maximum height. They express uncertainty about the use of equations and the relationship between velocity and acceleration due to gravity.
  • Some participants confirm the original poster's understanding of horizontal velocity remaining constant and clarify the concept of vertical velocity being zero at maximum height while still experiencing gravitational acceleration.
  • Further questions arise regarding the necessity of using vertical components to calculate time at maximum height, leading to discussions about the definitions of velocity and acceleration in this context.

Discussion Status

The discussion is ongoing, with participants providing clarifications and affirmations of the original poster's calculations. There is a productive exchange of ideas regarding the relationship between vertical and horizontal motion, though some confusion remains about the application of equations in different contexts.

Contextual Notes

Participants are navigating the complexities of projectile motion, particularly the distinction between vertical and horizontal components of motion. There is an emphasis on understanding the implications of zero vertical velocity at maximum height while still considering gravitational effects.

science_rules
Messages
157
Reaction score
2

Homework Statement


You throw a ball. The ball leaves your hand with a velocity of 15 m/s at an angle 60 degrees above the horizontal. Acceleration due to gravity is 9.8 m/s/s.
a) Determine the velocity (magnitude and direction) of the ball at the maximum height.
b) How much time does it take for the ball to reach maximum height?
c) Determine the maximum height.

Homework Equations


(I know how to draw the diagram, so that part is good)
I think the components of the velocity at maximum height can be found using these equations:
a) VoY= Vosin60 = 15m/s(sin60) = 12.99 m/s velocity of ball in vertical direction (initial)
VoX= Vocos60 = 15 m/s(sin60) = 7.5 m/s velocity of ball in horizontal direction (initial)
Since there is no vertical velocity at max height, then the velocity at max height is just the magnitude of the horizontal component: 7.5 m/s horizontal direction?

b) Time = Vy= VoY + ayt , set Vy = 0 at Y max.
0 = 12 m/s + (-9.8 m/s/s)t
t = (Vfy - Voy)/ -9.8m/s/s = (0 - 12.99)/-9.8 m/s/s = 1.32 seconds

c) Ymax. = Voyt + 1/2at^2 = (12.99 m/s)(1.32 sec.) + 1/2(-9.8m/s/s)(1.32 sec.)^2

= 8.60 meters max height


The Attempt at a Solution


see above. I am not sure if i did this problem correctly, or if i understand it quite. there are some things about it that i understand, others not so much.
I think the velocity at max height would just be the component 7.5 m/s in the horizontal direction, since suposedly, there is no vertical velocity at max height. Is this assumption correct? Have i used the correct equations and in the right context?? If not, please help me to understand this type of problem better.
I understand that at max height, there is only the force of gravity 9.8 m/s/s downward, the horizontal velocity for a projectile is constant, and there is no horizontal acceleration. I assume there is no horizontal acceleration, because the horizontal veloc. remains constant?
However, if there is no vertical velocity at max height then why is the object still accelerating at 9.8 m/s/s downward due to gravity? It seems like a contradiction, something can't be accelerating due to some force and not also have a velocity. Increasing velocity is the definition of acceleration, so there would be vertical velocity at max height. There is always vertical velocity due to the force of gravity accelerating the object downward.
 
Physics news on Phys.org
I think the velocity at max height would just be the component 7.5 m/s in the horizontal direction

Correct. The magnitude of the the two velocity vectors are as follows

sqr[ (7.5)^2 + (0)^2 ] = 7.5

Everything in your solution looks fine to me.

However, if there is no vertical velocity at max height then why is the object still accelerating at 9.8 m/s/s downward due to gravity?

Because firstly the object the object is moving in the positive direction but accelerates towards the negative direction. Eventually its going to have to cross the point of 0m/s in order to start moving in the negative direction instead of the positive direction.

Think of a thermometer, If its 20 degrees outside and it drops to -10 degrees it eventually crossed 0 degrees right?
 
Last edited:
i understand. thanks!
 
quick question, just to make sure i get this: the velocity for the ball at max height is the initial velocity component equation because the velocity for horizontal motion remains constant, am i right?? And the initial vertical velocity component does not apply to the first question (only with regard to finding time at max height using this equation Vy = Voy + ayt)
When i had first done this problem, i had confused the two initial component equations, not realizing this. i had tried to incorporate the vertical veloc. component equation with the first question even though i know that there is no vertical velocity. anyhow, now i know that i should have written the vertical component equation down near the second question(in my homework paper) to not confuse myself with the horizontal component equation. Both of them are found using a very similar equation. Do you see what i mean how I would confuse the two??
 
I think i had confused the two equations because i didn't understand why the time equation
t = Vfy - Voy / -9.8 m/s/s ) uses the vertical component instead of the horizontal component. is there a reason for this??
 
In other words, if there is no vertical velocity, why must i use a vertical veloc. component to find the time at max height. If vertical velocity is non-existent, why not use the horizontal veloc. component to find the time instead??
 
You're thinking that there is NO vertical component of velocity and therefore its impossible to solve for time when there is NOTHING to work with right?

You need to understand when we say there is no vertical component of velocity we mean Vy = 0.

And with that fact we are able to solve for time.
 

Similar threads

Replies
11
Views
2K
Replies
40
Views
3K
  • · Replies 38 ·
2
Replies
38
Views
4K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
18
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K