a) Let's first find the time that projectile takes to reach the highest point:
vy=v0sinθ−gt,0=v0sinθ−gt,t=gv0sinθ.The equation for the motion of projectile in the vertical direction can be written as follows:
y=v0sinθ−21gt2.Substituting t into the equation we can find the highest point reached by the projectile:
ymax=2gv02sin2θ,ymax=2⋅9.8 s2m(30 sm)2sin260∘=34.44 m.b)-c) Let's first find the time taken for flight:
tflight=2t=g2v0sinθ,tflight=9.8 s2m2⋅30 sm⋅sin60∘=5.3 s.Then, we can find the range of the projectile:
x=v0tflightcosθ,x=30 sm⋅5.3 s⋅cos60∘=79.5 m.(d) At the point when the path makes an angle of 30∘ to the horizontal we have:
tan30∘=vxvy=v0cosθv02sin2θ−2gh,h=2gv02sin2θ−v0cosθtan30∘,h=2⋅9.8 s2m(30 sm)2sin260∘−30 sm⋅cos60∘⋅tan30∘=34 m.
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