A 2.5kg cannonball is fired from the top of a cliff. The cannonball leaves the c

Question

A 2.5kg cannonball is fired from the top of a cliff. The cannonball leaves the cannon at a height of h=75 m above the ground below. The cannonball is fired at an initial speed of 120 m/s at an angle of =30 degrees above the horizontal. Use the coordinate system of up being the +y and right being the +x direction to calculate the following quantities for the motion of the cannonball. a)what is the maximum height the object reaches? b)when does the object reach its maximum height? c)what is the minimum speed of the cannonball? d)when does the cannonball return to a height of 75m after it has been fired? e) when does the cannonball hit the ground? f)what is the speed of the cannonball just before it hits the ground?

I understand this question is complex but I need to understand how to work this for my next exam so any help is much appriciated, thank you.

Explanation / Answer

here since there in no force along x axis (horizontal ) so velocity of the cannon ball remain same in x direction and acceleration due to gravity ,g(9.8m/s^2 ) will be in y-axis. initial velocity U(x) =120cos30 =103.92 m/s along y-direction u =120sin30 =60 m/s when it reaches at its maximum height its velocity(vertical) becomes zero v=u+at 0=60+(-9.8)t t =6.122 s (time to reach maximum height) height h=ut+(0.5)at^2 h=60*6.122+0.5*(-9.8)*(6.122)^2 =183.6734 m so maximum height =75+183.6734 =258.67 m time taken by the cannonball to return to the same level(75m) =2t=2*6.122 =12.245 s so let v is the velocity of the cannonball when it hits the ground 2as=v^2-u^2 2*9.8*258.67=V^2 v =71.20 m/s time taken to ground from maximum height v=u+at 71.20 =0+9.8*t t=7.266 seconds speed when it hits the ground =sqrt(103.92^2+71.20^2) =125.97 m/s answers a)maximum height =258.67 m (above ground) b)time when object reaches maximum height ,t=6.122 seconds c) minimum speed will be when vertical component of the velocity becomes zero i.e at maximum height minimum speed = Ux =103.92 m/s d) time taken by the cannonball to reach 75 m =12.245 seconds e) time taken by the cannonball to hit the ground =6.122+7.266=13.388 seconds f) speed of the cannonball just before it hits the ground =125.97 m/s

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