How Long Does an Apple Stay in the Air When Thrown Upwards?

  • Thread starter Thread starter doctordiddy
  • Start date Start date
  • Tags Tags
    Air apple Time
AI Thread Summary
The discussion focuses on calculating the total time an apple stays in the air when thrown upwards from a height of 1.3 meters with an initial velocity of 2.7 m/s. The maximum height reached is approximately 1.67156 meters, and the equations of motion are applied to find the time taken to reach this height and the subsequent fall. A quadratic equation arises from the motion equations, but there is confusion about isolating the variable t due to the presence of both t and t^2. Participants clarify that the initial conditions must be correctly interpreted, and the quadratic formula is suggested as a solution method. Ultimately, the discussion emphasizes the importance of accurately defining the problem to find a valid solution.
doctordiddy
Messages
53
Reaction score
0

Homework Statement



Basically a girl throws an apple vertically upward from a height of 1.3m with an initial velocity of 2.7m/s. I also know that gravity is =-9.8m/s. I found that the apple reaches a maximum height of 1.67156 m. How long does the apple stay in the air in total?

All I know is I have to find the time it takes is made up of the apple being thrown up, and then falling down all the way past the starting point to the ground. (the starting point is 1.3m off the ground)



Homework Equations



I think you have to use d=vit + 1/2at^2 and maybe vf^2 = vi^2 + 2ad

d= distance
a= acceleration
t= time

The Attempt at a Solution



I know the first equation to find the time it takes the apple to reach max height but after that I just don't know how to solve it.

I end up with an equation with two t's, one t and one is t^2

Here is my work for finding t1

1.67156 = 2.7t - 4.9t^2

How do i isolate t if there are two t's?
 
Physics news on Phys.org
doctordiddy said:


1.67156 = 2.7t - 4.9t^2

How do i isolate t if there are two t's?


To solve, you need to use the quadratic formula.

I think there is an error in your thinking; you are finding the time to travel up 1.67 m starting at 2.7 m/s against gravity. This isn't possible - there is no solution (try it).

You are on the right track. How high is the ball when it is starts (at Vi = 2.7 m/s) and where is the ground relative to the start position?
 
Last edited:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top