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

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SUMMARY

The discussion centers on calculating the total time an apple remains in the air when thrown vertically from a height of 1.3 meters with an initial velocity of 2.7 m/s, under the influence of gravity at -9.8 m/s². The maximum height reached by the apple is determined to be 1.67156 meters. To find the total time in the air, the quadratic formula is necessary due to the presence of two time variables in the equation derived from the motion equations: d = vit + 1/2at² and vf² = vi² + 2ad.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = vit + 1/2at²
  • Familiarity with the quadratic formula for solving equations
  • Knowledge of gravitational acceleration (-9.8 m/s²)
  • Basic principles of projectile motion
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  • Learn how to derive maximum height in projectile motion problems
  • Explore the concept of initial and final velocity in vertical motion
  • Investigate the effects of gravity on objects in free fall
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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?
 
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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?
 
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