Calculus
posted by Desperate!!!! on .
A blast blows a rock straight up at 150 ft/sec. The path of the rock is described by the s(t) = 150t16t^2 where s is measured in feet and t in seconds. Find the velocity of the rock when it is 200 ft above the ground and when the rock hits the ground.

v(t)=s'(t)=15032t
set s(t)=200 find out when the rock is 200 feet above the ground.
200=150t16t^2
t=(5/16)+or(15sqrt(97))
0=150t16t^2
t=75/8 not 0 because it started on the ground
the velocity when it strikes the ground is v(75/8)=150
the other velocity is v((5/16)+or(15sqrt(97)))=304.387 or 24.836