Calculus Help??

Question

If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 80 ft/sec, its height after t seconds is s(t) = 32 + 80t - 16t^2 .

What is the maximum height the ball reaches?

What is the velocity of the ball when it hits the ground (height = 0 )?

Answer 1

hmax is for t = 80/32
so hmax = 32+6400/32-16*6400/32^2=132 feet
h=0 for -16t^2+80t+32=0
t^2-5t-2=0 t =( 5+-sqrt(33))/2 = 2.5+2.87 = 5.37 s
v= -32t+80=-91.84 m/s(- means downward)

Answer 2

SInce s(t) is position, take the first derivastive and set it to zero. That will given the time of teh minimum or maximum of the function s(t).

ds/dt = 0 = 80 -32t ---> 80 = 32 t or t =80/32 = 5/2 = 2.5 seconds

so max height occurs at t = 2.5 sec and

s(2.5) = sm = 32 +80*2.5 - 16*(2.5)^2

Speed is ds/dt = 80 - 32 t so for the second part you need to find t such that s(t) = 0.


0 = 32 +80t -16t^2

0 = 4 + 10t - 2t^2

0 = 2 +5t - t^2 = t^2 -5t - 2

t =5/2+/- 1/2*sqrt(5^2+4*2)

Choose + sing and substitute into ds/dt