will stand by to award points and ask questions On level ground a shell is fired

Question

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On level ground a shell is fired with an initial velocity of 85.0m/s at 63.2? above the horizontal and feels no appreciable air resistance.

part a Find the horizontal and vertical components of the shell's initial velocity.

part b How long does it take the shell to reach its highest point?

part c Find its maximum height above the ground.

part d How far from its firing point does the shell land?

part e At its highest point, find the horizontal and vertical components of its acceleration

part f At its highest point, find the horizontal and vertical components of its velocity.

Explanation / Answer

part a
horizontal component ux= 85cos63.2=38.32 m/s
vertical component uy = 85sin63.2= 75.87 m/s
part b
at highest point vy=uy+ayt=0
or 75.87-9.8t=0
or t=7.74 sec

part c
since distance s= ut+(1/2)at2
so max hieght at t= 7.74 sec
  s= [75.87*7.74]-[(1/2)9.8*7.742]
or s= 293.69 m

part d
since total time of journy will be double of total time at highest point.
so T =2t
T=2*7.74=15.48 sec
so horizontal full diatance = 15.48*38.32 =593.19m {since horizontal accelaration is zero}

part e
since horizontal accelaration is zero
so at max height its component will be zero.
and vertical component of accelaration at highest point will be 9.8 m/sec2......direction will be downward to ground.

part f
since horizontal velocity is constant so it at highest point horizontal component of velocity will be 38.32 m/s
and vertical component will be zero since at highest point velocity dowards vertical will be zero.

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