Using calculus, it can be shown that if a ball is thrown upward with an initial velocity of 32 ft/s from the top of a building 594 ft high, then its height h above the ground t seconds later will be
h = 594 + 32t − 16t2.During what time interval will the ball be at least 34 ft above the ground?

Question

Using calculus, it can be shown that if a ball is thrown upward with an initial velocity of 32 ft/s from the top of a building 594 ft high, then its height h above the ground t seconds later will be 
h = 594 + 32t − 16t2.During what time interval will the ball be at least 34 ft above the ground?

anonymous · Answer

Solve 
594 + 32 t - 16 t^2=34 for t

anonymous · Answer

you get t=-5 and t=7
between 0 and 7

anonymous · Answer

$a(t)=-32$ as gravity sucks $v(t)=\int a(t)dt =-32t+v_0$ and since you are told $v_0=32$ we get $v(t)=-32t+32$
then $h(t)=\int v(t)dt=-16t^2+32t+s(0)$ and since $h(0)=594$ we have 
$$h(t)=-16t^2+32t+594$$ 
i wish i could learn to read
i thought it said "show" not "it can be shown"
what a waste

anonymous · Answer

But not reading, you gave a review of how to get it.