A 5 kg object is launched from a 35 m high cliff at an angle of 30 degrees with a speed of 100 m/s. How far from the cliff does it hit the ground

The object will hit the ground at 882.77 m from the lunch

Explanation:

we are to find the range of the motion

What is range?

It is the distance from the point of projection to the point where the object hits the ground

R=(v^2sin2θ)/g

Given data

v= 100 m/s

θ= 30 degree

g= 9.81 m/s^2

Substituting to find the R we have

R= 100^2 *sin(2*30)/9.81

R= (10000*0.866)9.81

R= 8660/9.81

R= 882.77 m

the answer would be a

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