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

##### 1 Answer

1.275 m

#### Explanation:

First, break the initial velocity of the object into

#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

#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

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

#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#