# A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #12 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

##### 1 Answer

#### Explanation:

As you know, a projectile launched from the ground at an angle **not** equal to *straight up*, will have a motion that can be broken down into a **vertical component** and a **horizontal component**.

**Vertically**, the projectile will be acted upon by *gravity*. This means that its vertical movement can be broken down into two parts

*from***ground level**to**maximum height***from***maximum height**to**ground level**

On its way up, the projectile is **decelerated** by the *gravitational acceleration*, **accelerated** by the same gravitational acceleration.

The key here is to realize that **at maximum height**, the **vertical component** of its velocity will be **equal to zero**.

**Horizontally**, on the other hand, the projectile's motion in unaffected by gravity, or by any other force for that matter. This implies that the **horizontal component** of its velocity will be **constant** throughout its motion.

So, if we take

#v_(0v) = v_0 * sin(theta) -># thevertical componentof its initial velocity

#v_(0h) = v_0 * cos(theta) -># thehorizontal componentof its initial velocity

Your strategy now will be to find the **time** needed for the projectile to reach maximum height, then use that time to calculate the horizontal distance it covered.

So, you know that at *maximum height* you have

#v_v = 0#

This means that you can write, for the vertical component of the projectile's motion,

#overbrace(v_v)^(color(purple)(=0)) = v_(0v) - g * t" "# , where

*maximum height*

Rearrange to solve for

#t = v_(0v)/(g)#

Plug in your values to get

#t = (12color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s"^(-1)))) * sin((5pi)/12))/(9.81 color(red)(cancel(color(black)("m")))"s"^color(red)(cancel(color(black)(-2)))) = "1.18 s"#

This means that you can write, for the horizontal component of the motion

#d = v_(oh) * t#

#d = 12 * cos((5pi)/12)"m" color(red)(cancel(color(black)("s"^(-1)))) * 1.18color(red)(cancel(color(black)("s"))) = color(green)("3.66 m")#