(A)
We know that
gu2sin(2θ)=range=700putting all values9.86302sin(2θ)=700sin(2θ)=0.0172839
θ=0.5°tan(θ)=0.0087269=700xx=700×0.0087269=6.1 m
So he should aim 6.1 m above the targe)
(B) max height = 2gu2(sin(θ))2=2×9.86302×(0.0087265)2=1.54207 m
(C) Time of flight = g2usin(θ)=9.82×630×0.0087265=1.122 sec
Comments