Projectile Motion- Object Launched in the Air

1. Apr 19, 2017

Catchingupquickly

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

2. Relevant equations
$\vec v_1v = \vec v_1 sin\Theta$

$\Delta \vec d = \vec v_1v \Delta t + \frac 1 2 \vec a \Delta t^2$

3. 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?

2. Apr 19, 2017

scottdave

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

3. Apr 19, 2017

Catchingupquickly

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

4. Apr 19, 2017

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?