Question

A projectile is launched into the air. The function h(t) = –16t2 + 32t + 128 h hieght t seconds After how many seconds will the projectile land back on the ground?

Answer 1

-16t^2 + 32t + 128 = 0

Dividing the equation by -16 to simplify, we get:

t^2 - 2t - 8 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation t^2 - 2t - 8 = 0, the coefficients are:

a = 1, b = -2, and c = -8.

Substituting these values into the quadratic formula, we get:

t = (-(-2) ± √((-2)^2 - 4(1)(-8))) / (2(1))

Simplifying further:

t = (2 ± √(4 + 32)) / 2

t = (2 ± √36) / 2

t = (2 ± 6) / 2

This gives us two possible values for t:

t1 = (2 + 6) / 2 = 8 / 2 = 4

t2 = (2 - 6) / 2 = -4 / 2 = -2

Since time cannot be negative in this context, we discard t2 = -2. which, the projectile will land back on the ground after 4 seconds.