A plane is flying horizontally at 98m/s and releases and object which reaches the ground in 10s. The angle made by it while hitting th ground is:
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The angle can be find just by finding the vertical component and horizontal component of velocity with which it will hit the ground.
So,considering for vertical motion,velocity after 10s will be,
v=0+gt (as,initially downward component of velocity was zero)
so,v=9.8⋅10=98ms−1
Now,horizontal component of velocity remains constant through out the motion i.e 98ms−1 (as because thisvelocity was imparted to the object while releasing from the plane moving with this amount of velocity)
So,angle made with the ground while hitting is tan−1(9898)=45∘
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