## Time an apple stays in the air

1. The problem statement, all variables and given/known data

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)

2. Relevant 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

3. 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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 Quote by doctordiddy [b] 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?