If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 16 ft/sec, its height after t seconds is s(t)=64+32t−16t2. what is the velocity of the ball when it hits the ground (height 0)?

Question

campbell_st · Answer

ok... you need to find the time when the ball hits the ground.... which is when the height 

s(t) = 0

so you need to solve 

$$64 + 32t - 16t^2 = 0$$

which is 

$$16(4 + 2t - t^2) = 0$$

the quadratic needs toe general quadratic fromula to find the time when the ball hits the ground. 

next. 

differentiating s(t) will give the velocity equation. 

when that is done substitute the value of t when the height = 0

this will give the velocity

campbell_st · Answer

oops forgot to say... when you find t only use the positive value.... its a bit over 3...

anonymous · Answer

so $$-t^2+t+2=0$$ then factor out, to find the positive number

campbell_st · Answer

nope... 

if you take 16 as a common factor you are having to solve 

$$-t^2 + 2t + 4 = 0$$

anonymous · Answer

i did the quadratic equation earlier and got a number +3.23 if that is what you meant? i plugged that in or just 3 into the equation but didnt seem to get the answer

campbell_st · Answer

ok... thats good

now can you find the derivative of 

$$s(t) = -16t^2 + 32t + 64$$

campbell_st · Answer

plugging your value of t into the displacement equation s(t) will give 0..