Problema Solution
Mike threw a ball in the air. The height of the ball is given by the equation h=-5t^2+15t+140 where h is height in meters of the ball above the ground t seconds after it is thrown. How many seconds after Mike throws the ball will it hit the ground?
Answer provided by our tutors
Mike hits a golf ball into the air. The equation that describs the path of the ball is h=55t-5t^2. There are three questions.
(A How high is it the ball after 2 seconds?)
h=55t-5t^2
Simply substitute 't' with 2 and solve:
h=55(2)-5(2)^2
h=110-5(4)
h=110-20
h=90 meters
.
(B When will it first reach 140 meters?)
h=55t-5t^2
Subsitute 'h' with 140 meters and solve:
140=55t-5t^2
0=55t-5t^2-140
0=-5t^2+55t-140
0=-t^2+11t-28
0=t^2-11t+28
0=(t-4)(t-7)
t = {4,7}
Ball reaches height at 4 and 7 seconds
.
(C When will it hit the ground?)
h=55t-5t^2
Subsitute 'h' with 0 meters and solve:
0=55t-5t^2
0=11t-t^2
0=-t^2+11t
0=t^2-11t
0=t(t-11)
t = {0,11}
Ball is at 0 meters at 0 seconds and 11 seconds