How do I solve for Δt when there's a Δt^2
Just use the quadratic formula.
So what would b, a and c be?
when you set the equation equal to zero a is the t^2 value, b is value with t variable, and c is just the number without the variable with it.
"ax^2 + bx + c"
I thought you were trying to solve for the change in time? So you would set the equation equal to zero d = d∅ + v∅t + 1/2 vt^2
and you just put d - d∅ = Δd
Δd = v∅t + 1/2 vt^2
where v∅=initial velocity
so the v∅t would be the b value
1/2 * v would be the a value
you would subtract Δd from both sides to get the equation equal to 0 and Δd would be your c value
it would look like
0 = 1/2vt^ + v∅t - Δd
0 = at^2 + bt + c
so you'll get two answers one at the max height of trajectory when velocity = 0 and one at the end of the trajectory where velocity = 0 again
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