Question

A projectile is shot at an angle of `(5pi)/12` and a velocity of `5 m/s`. How far away will the projectile land?

Answer 1

1.275 m

Explanation:

First, break the initial velocity of the object into `x` and `y` components.

`v_x = 5 cos((5pi)/12) m/s = 1.294 m/s`

`v_y = 5 sin((5pi)/12) m/s = 4.830 m/s`

The formula for figuring out how far a projectile will fly is:

`d = "fall time" times v_x`

The fall time can be found using the kinematics equation:

`y = y_0+v_yt + 1/2at^2`

In this case, we know `y_0 = 0` `v_y = 4.830` and `a = g = -9.8 m/s^2`, and we're trying to find the time when `y=0` (when it hits the ground again).

`0 = 4.830t - 4.9t^2`

`0 = t(4.830-4.9t)`

`t = 0 and 4.9t = 4.830`

`t = 0 and t = 0.9857`

So the object has a fall time of `0.9857` seconds.

Now all we have to do is use the projectile distance formula:

`d = "fall time" times v_x`

`d = 0.9857 * 1.294 = 1.275 m`

Final Answer

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As a side note for the future, when the starting height of an object is `0`, you can derive a formula for projectile distance which will save time:

`y = y_0+v_yt+1/2at^2`

`0 = vtsintheta-g/2t^2`

`0 = t(vsintheta-g/2t)`

`0 = vsintheta-g/2t`

`g/2t = vsintheta`

`t = (2vsintheta)/g`

Plugging in this value for the fall time, we get:

`d = "fall time" times v_x`

`d = (2vsintheta)/g times vcostheta`

`d = (v^2(2sinthetacostheta))/g`

`d = (v^2sin(2theta))/g`