Projectile Motion: An Egg Thrown Vertically Upwards

In summary, the egg will be located 25.0 m above the ground at the following two time values: 2.90s and 1.07s.
  • #1
TheFlemster
25
0
Homework posted in wrong forum, so no template
1- A person throws an egg directly upward with an initial speed of 27.4 m/s. When the person

throws the egg it is located 1.36 m above the ground at the exact moment of release. The effects of air-resistance on the egg can be ignored. (Take the moment of release as t = 0).

a. What is the egg's maximum height above the ground?
b. At what two time values will the egg be located 25.0 m above the ground?

I got part a as 39.7 m. And the upward time of part b as 1.07s. I can not figure out the downward time
 
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  • #2
How did you get part (a)? What equation(s) did you use?
 
  • #3
Moderator note: Title changed to reflect actual thread content. Note to @TheFlemster: Pleas for help are not permitted as thread titles. See the forums rules or the pinned thread, Guidelines for students and helpers .
 
  • #4
We know at its highest point, V is equal to zero. I used v^2 = Vo^2 + 2a • (X - Xo)
Where v is final velocity. Vo is initial velocity. a is acceleration due to gravity. X is final displacement Xo is initial displacement
 
  • #5
OK, that's good. Is there a kinematics equation (other than the one you already used) that you can use to describe the height of the egg above ground at any time t?
 
  • #6
X = Xo + Vot + 1/2at^2
Where t is time
 
  • #7
That's the one. Can you take this general form and adapt to the specifics of your problem?
 
  • #8
I tried. I get X - Xo = t(Vo + 1/2at)
I couldn't get anywhere with it. The answers were 1.07s and 2.90s. We were given answers but not how to get to the answer
 
  • #9
Just a moment, don't give up so soon. Answer the following questions first and you will find your way.
There are 5 symbols in the equation. What does each symbol stand for? Describe with words.
Which of these symbols corresponds to numbers that are given to you?
 
  • #10
X is the final displacement of 39.7 m
Xo is initial distance of 1.36m
T is time
Vo is initial velocity of 27.4m/s
a is acceleration due to gravity at -9.8m/s^2
 
  • #11
In post #5 I asked you to write an equation that gives the height of the egg above ground at any time t. This means that you need to express one variable, namely "position above ground" as a function of another variable, namely "any time t". In other words, write an equation such that if I give you any time I please, you can give me the height of the egg above ground at that time. Can you do that? Hint: The 39.7 m is the height at a specific time, not at any time.
 
  • #12
Image1486668835.472271.jpg


This is what I got
 
  • #13
You are going around in circles. Let's start with the equation that I asked you to analyze. I will give you wht the symbols stand for and you will have to put it together.
x = x0 + v0t + ½ a t2
The following have the following specific and reserved meanings
x = the position of the object at any time t, a.k.a. the dependent variable often written as x(t)
x0 = the position of the object at the specific time t = 0
v0 = the velocity of the object at the specific time t = 0
t = placeholder for any time t, a.k.a. the independent variable
a = the constant acceleration

How would you go about adapting this equation to the problem you have? Start by identifying which of these quantities are given to you. The 39.7 m you got in part (a) is irrelevant for part (b) and out of the picture. Once you complete the adaptation, you will have an equation that gives the height of the egg above ground at any time t. Get that first, and we will worry about what to do with it later.
 
  • #14
TheFlemster said:
X = Xo + Vot + 1/2at^2
Where t is time
What happens if you put ##X = 25m## in this equation?
 

Related to Projectile Motion: An Egg Thrown Vertically Upwards

1. What is projectile motion?

Projectile motion is the motion of an object through the air that is subject to only the force of gravity. It follows a curved path known as a parabola.

2. How does the shape of the path in projectile motion change?

The shape of the path in projectile motion depends on the initial velocity and angle at which the object is thrown. A higher initial velocity or a steeper angle will result in a longer and narrower parabola, while a lower initial velocity or a smaller angle will result in a shorter and wider parabola.

3. What is the effect of air resistance on projectile motion?

Air resistance can affect the motion of a projectile by slowing it down and changing its trajectory. This is because air resistance creates a force in the opposite direction of the object's motion, causing it to lose speed and deviate from its original path.

4. In what ways does the launch angle affect the maximum height of a projectile?

The launch angle of a projectile has a direct impact on its maximum height. A launch angle of 90 degrees (straight up) will result in the maximum height, while any other angle will result in a lower maximum height. The steeper the angle, the lower the maximum height will be.

5. How can the equations of motion be used to calculate the height and velocity of a projectile at any given time?

The equations of motion, specifically the equations for displacement and velocity, can be used to calculate the height and velocity of a projectile at any given time. By plugging in the initial conditions (such as launch angle and initial velocity) and the time, the equations can be solved to determine the height and velocity at that specific time during the projectile's flight.

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