In the figure, a stone is projected at a cliff of height *h* with an initial speed of 42.0 m/s directed at angle θ_{0} = 60.0° above the horizontal. The stone strikes at *C*, 5.50 s after launching. Find

(a) the height *h* of the cliff,

(b) the speed of the stone just before impact at *C*, and

(c) the maximum height *H* reached above the ground.

In this problem, we’re asked to calculate the **height of the cliff** the stone lands on, the **final speed **of the stone, and the **maximum height** the stone reaches during its motion.

This is a __projectile motion__ problem with a positive launch angle. For projectile motion problems, we follow the following simple steps:

- Draw diagram, axes.
- Develop equations to describe the various intervals.
- Solve each of the target variables.

**Step 1: **Draw diagram, axes.

For projectiles with a positive launch angle, the motion has two parts: **AB**, from launch to peak, and **BC**, from peak to when it lands. We use **AC** for the full motion.

We'll consider the x and y speed components separately. Therefore, to resolve v_{0} to v_{0x} and v_{0y}, we apply trigonometry

${\mathbf{v}}_{\mathbf{x}}\mathbf{=}{\mathbf{v}}_{\mathbf{0}\mathit{x}}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}\mathbf{Cos\theta}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{42}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{Cos}\mathbf{}\mathbf{(}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{)}$

v_{x} = **21.0 m/s**

The horizontal speed (x-component of v_{0}) is constant since there is no horizontal acceleration.

${\mathit{v}}_{\mathbf{o}\mathbf{y}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{v}}_{\mathbf{0}}\mathit{S}\mathit{i}\mathit{n}\mathit{\theta}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{42}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{)}\mathbf{}\mathit{S}\mathit{i}\mathit{n}\mathbf{(}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{\xb0}\mathbf{)}$

v_{0y} = **36.4 m/s**

The acceleration of the stone before reaching maximum height causes a decrease in vertical speed, v_{y} while the acceleration after the maximum height increases the vertical speed, v_{y},_{ }downwards.

We will also use a **standard coordinate system** where __upward__ motion is **positive y** and __downward __motion is **negative y**. Also, the stone moves to the __right__ in the **positive** **x-direction**.

Projectile Motion: Positive Launch

Projectile Motion: Positive Launch

Projectile Motion: Positive Launch

Projectile Motion: Positive Launch