# A projectile is shot from the ground at a velocity of #17 m/s# and at an angle of #(7pi)/12#. How long will it take for the projectile to land?

##### 1 Answer

#### Answer:

#### Explanation:

We're asked to find the time when the projectile lands after being launched with a known initial launch velocity (magnitude *and* direction. In terms of physics, we need to find the time *height*) is equal to zero.

To solve this, we can use the equation

Knowing that

Therefore, plugging in known variables, we have

rounded to

Therefore, the object will land after

Something we could do (for future reference and for fun) is to find

a general expression for the time *any* launched projectile when its height is the same as its launch height (zero, in this case).

Let's write our original equation but leave all the variables un-substituted, except for

What we can first do is divide both sides by

Now, let's move the acceleration group to the left side:

Multiplying both sides by

And isolating

Thus, **whenever a projectile is launched from any height, the time #t# when it reaches that height again is given by the equation #t = (2v_(0y))/g#**

Intuitively, the projectile will only reach the same height again if it is launched (at least somewhat) upward; that is, it's initial

We can check this from the equation: if *even before it was launched* (the motion equation, if graphed, will show that this height is its *maximum height*; that is, if the projectile was launched at a time *before* it was in your hand, it would reach your hand at this maximum height).

If *negative*, it was launched (at least somewhat) downward, and the time given will be negative, which makes no sense *realistically*, but *theoretically*, if it continued the motion by which it was projected *before* it was launched, it would be at that height at that calculated time, before it was even launched.

Lastly, let's check to see if this general equation works (if it gives us the same time):

Indeed it does:)