Projectile Motion- Object Launched in the Air

[Catchingupquickly]

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?

 

[scottdave]

Everything looks good to me. Personally, I keep the intermediate answers, in my calculator, to full digits, then round at the end.

 

[Catchingupquickly]

Thank you. Yeah, overall, the textbook isn't the greatest, but I'm just following its lead for this course.

 

[haruspex]

You were not asked for the time. You can get to the height directly by using the right SUVAT formula. Which one does not involve time?