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Catchingupquickly
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Homework Statement
A spring toy is launched from the ground at 2.3 m/s at an angle of 78° to the ground.
What is the maximum height reached by the spring toy?
Homework Equations
## \vec v_1v = \vec v_1 sin\Theta##
## \Delta \vec d = \vec v_1v \Delta t + \frac 1 2 \vec a \Delta t^2##
The Attempt at a Solution
## \vec v_1v = 2.3 m/s [up] (sin78 degrees)
\\ = 2.25 m/s [up]##
## \Delta \vec d = \vec v_v1 \Delta t + \frac 1 2 \vec a \Delta t^2
\\ 0 = (2.25 m/s [up]) \Delta t + \frac 1 2 (-9.8 m/s^2) \Delta t^2
\\ 0 = \Delta t (2.25 m/s - 4.9 m/s^2) \Delta t
\\ \Delta t = 0.46 seconds##
That's the total time it was airborne. Maximum height is half that so 0.46/2 = 0.23 s.
Now to find the height.
## \Delta \vec v = \vec v_1 \Delta t + \frac 1 2 \vec a \Delta t^2
\\ = (2.25 m/s) (0.23 s) + \frac 1 2 (-9.8 m/s^2) (0.23)^2
\\ = 0.52 - 0.26
\\= 0.26 m [up]##
The max height is 0.26 meters [up].
I've followed my textbook's lead of rounding to two significant places through the work.
Am I correct at all?