Asked by Desperate!!!!
A blast blows a rock straight up at 150 ft/sec. The path of the rock is described by the s(t) = 150t-16t^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.
Answers
Answered by
john
v(t)=s'(t)=150-32t
set s(t)=200 find out when the rock is 200 feet above the ground.
200=150t-16t^2
t=(5/16)+or-(15-sqrt(97))
0=150t-16t^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-(15-sqrt(97)))=304.387 or -24.836
set s(t)=200 find out when the rock is 200 feet above the ground.
200=150t-16t^2
t=(5/16)+or-(15-sqrt(97))
0=150t-16t^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-(15-sqrt(97)))=304.387 or -24.836
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