- #1
ssjcory
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Homework Statement
An object is thrown downward with an initial speed of 10 m/s from a height of 60m above the ground. At the same instant, a second object is propelled vertically upward from ground level with an initial speed of 40 m/s. At what height above the ground will the two objects pass each other?
Homework Equations
Professor said for where+when problems should always use the formula
Xf = X0 + (VX0)(Δt) + (1/2)(AX0)(Δt)^2
Variable key:
Xf = final x
X0 = initial x
VX0 = initial velocity
AX0 = acceleration due to gravity (-9.8m/s^2)
Δt = change in time
The Attempt at a Solution
Ok so first I mapped out the known factors in the equation for each motion of interest
For motion A we have:
X0 = 60, VX0 = 10, AX0 = 9.8, missing Δt and Xf
For motion B we have:
X0 = 0, VX0 = 40, AX0 = 9.8, again missing Δt and Xf
I figure that we are looking for the X when both x's are zero so I set them to equal each other in the same form as the equation
AX0 + AVX0(Δt) + (1/2)(AX0)(Δt)^2 = BX0 + BVX0(Δt) + (1/2)(AX0)(Δt)^2
60 + 10t + 4.9t^2 = 40t - 4.9t^2
From this I get everything to one side so the right side is 0 and I can use the quadratic formula to solve for Δt
Equation looks like
9.8t^2 - 30t + 60 = 0
So i do the quadratic equation
30 +/- sqrt(30^2 - 4(9.8)(60)) all over 2(9.8)
I end up with 3.47475 OR -.413526 as Δt.
I plug either one of those numbers into either one of the original equations to solve for Xf and I get numbers that are no where near correct.
He has listed the answer as 41 m.
Am I going about this completely wrong?
Thanks,
Cory