Question

A projectile is shot from the ground at a velocity of `4 m/s` and at an angle of `(5pi)/6`. How long will it take for the projectile to land?

Answer 1

`t = 0.408` `"s"`

Explanation:

We're asked to find the time `t` it takes a projectile to land, given its initial velocity.

When the particle lands, its height will be `0` (assuming it launches and lands at the same height), and we can use the kinematics equation

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

where

• `Deltay` is the height (`0`)

• `v_(0y)` is the initial `y`-velocity, equal to

`v_(0y) = v_0sinalpha_0 = (4color(white)(l)"m/s")sin((5pi)/6) = 2` `"m/s"`

• `t` is the time (what we're trying to find)

• `g` is the acceleration due to gravity near earth's surface, `9.81` `"m/s"^2`

Plugging in known values, we have

`0 = (2color(white)(l)"m/s")t - 1/2(9.81color(white)(l)"m/s"^2)t^2`

`1/2(9.81color(white)(l)"m/s"^2)t^2 = (2color(white)(l)"m/s")t`

`1/2(9.81color(white)(l)"m/s"^2)t = 2color(white)(l)"m/s"`

`t = (2color(white)(l)"m/s")/(1/2(9.81color(white)(l)"m/s"^2)) = color(red)(0.408` `color(red)("s"`