Question

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

Answer 1

`2.4` `"s"`

Explanation:

We're asked to find the time when the projectile lands when it is launched with a known initial velocity. We need to find the time `t` when the height `Deltay` is `0`.

We can use the equation

`Deltay = v_(0y)t - 1/2g t^2`

to find this. Our known quantities are

• `Deltay` must be `0`, because we're trying to find the time when this occurs.

• to find the initial `y`-velocity `v_(0y)`, we can use the equation `v_(0y) = v_ysinalpha`, so `v_(0y) = 12"m"/"s"sin((7pi)/12) = 11.6"m"/"s"`

• `g`, the acceleration due to gravity near Earth's surface, is `9.8"m"/("s"^2)`

If we make `Deltay` zero, we can rearrange the equation to solve for `t`:

`0 = v_(0y)t - 1/2g t^2`

`t = (2v_(0y))/g`

Therefore,

`t = (2(11.6cancel("m")/"s"))/(9.8cancel("m")/(cancel("s"^2))) = color(red)(2.4` `color(red)("s"`

The projectile will land after `2.4` seconds.