Baseball Problem: Does a Ball Hit the Roof? Answer Here!

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To determine if the baseball hits the roof of the domed stadium, the maximum height of the ball must be calculated. The ball is thrown vertically at an initial velocity of 110 ft/sec from a height of 6 feet. Using the equations of motion under gravity, the height as a function of time can be derived. The maximum height can be found by integrating the velocity function, which accounts for gravitational acceleration. If the calculated maximum height exceeds 200 feet, then the ball will hit the roof.
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Homework Statement


This is part of a calculus test and I am clearly missing something because this should be an easy problem: A baseball pitcher is having fun with the fans. He can throw a ball vertically froma point 6 feet above the ground at 75 miles per hour (110 ft/sec). He is standing in a domed stadium with a roof 200 feet above the ground. Does the ball hit the roof? Justify your answer.

We are currently doing integrals and just finished derivatives if that's any help. Any help would be appreciated, thanks.


The Attempt at a Solution


I sadly don't have one.
 
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Clearly you need to figure out the maximum height the ball reaches. Can you write an expression for the height as a function of time?

You can start with the acceleration due to gravity and integrate if you like.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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