h(t)=-16t^2+80t

Question

The equation h(t)=-16t^2+80t describes the height "h" of a ball thrown up in the air, where "t" is the time in seconds. How many seconds does it take the ball to reach the ground?

Answer 1

h(t) = -16t²+80t + 0  seems a bit off. This means we are releasing the ball from a height of zero feet.

 

(Doesn't seem possible)    Probably a problem more like h(t) = -16t2+80t + 6  (released from 6ft off the ground) makes more sense.

 

But let's go back to your problem.

 

We are solving for the t when h(t) = 0, therefore we will set our function -16t²+80t = 0

 

Since this is a quadratic with the C term missing,  (Ax²+Bx+C)

 

we will solve by factoring.

 

factor a t out 

 

t(-16t+80) = 0

 

Not set each factor = 0

 

the first piece is really easy t = 0

 

2nd piece -16t+80 = 0

 

Solve by isolating t

 

80 = 16t

 

divide both sides by 16

 

5 = t

 

So the ball is on the ground a time , t = 0 (right before you threw it) and goes up and comes back down and hits the ground 5 seconds later, t = 5

Answer 2

solve:

0=-16t^2+80t

t=5 seconds