Projectile motion: h = -1/2 gt^2 + vot + ho. A projectile has an initial velocity of 34.3 m/s and is launched 2.1 m above the ground.

Question

anonymous · Answer

$$h = -1/2 g t^{2} + v _{o}t + h _{o}$$

anonymous · Answer

What is the maximum height, in metres reached by the projectile?

anonymous · Answer

The maximum height is when V2 in the y direction = 0
If the initial velocity is all in the y direction (straight up), use
V2 = V1 +at      to find the time for it to reach the maximum height.

Then you can use the equation you wrote to find the height above the position it was projected from.

anonymous · Answer

The maximum height of the projectile is$$y_{\max} = {v^2 \sin^2(\theta) \over 2g}$$

anonymous · Answer

eashmore, it would be good to show where the equation came from. I assume you are using an energy equation, rather than projectile motion.
1/2mv^2 = mgh     because the initial height is 0, and the final velocity = 0

So, h = V^2/2g

Either way works.

anonymous · Answer

It comes from the basic kinematics of the problem. $$y = v \sin(\theta) t - {1 \over 2} g t^2$$It can be shown that time to reach maximum height is $$t = {v \sin(\theta) \over g}$$ since $v_y = 0$ at the top of the trajectory  $$v_y = v \sin(\theta) - g t \rightarrow t = {v \sin(\theta) \over t}$$Substituting the time to reach the max height into the expression for height, yields$$y = v \sin(\theta) \cdot \left( v \sin(\theta) \over g \right) - {1 \over2} g \cdot \left(v \sin(\theta) \over g \right) ^2$$Combining like terms$$y = {v^2 \sin^2 (\theta) \over g} - {1 \over 2} {v^2 \sin^2 (\theta) \over g}$$Performing the addition$$y_\max = {v^2 \sin^2(\theta) \over 2g}$$

anonymous · Answer

There is a slight mistake in the derivation of the time to reach the top of the trajectory, but the equation is listed correctly above as$$t = {v \sin(\theta) \over g}$$

anonymous · Answer

Additionally, as you suggested, we could find the maximum height from energy. Since all the vertical kinetic energy at launch will be converted to potential energy at the top of the trajectory. $${1 \over 2} m (v \sin(\theta))^2 = mgy_{\max}$$mass will cancel out. Solving for $y_{max}$$$y_{\max} = {v^2 \sin^2 (\theta) \over 2g}$$This is more efficient way to find the maximum height.