Question

An object is launched at 18.4 meters per second (m/s) from a 36.8-meter tall

platform. The equation for the object's height s at time t seconds after launch is
s(t) = -4.9t2 + 18.4t + 36.8, where s is in meters.

Answer 1

Explanation:

Given the equation for the object's height s at time t seconds after launch is

s(t) = -4.9t^2 + 18.4t + 36.8

Since are not asked what to look for, we can as well lok for the time taken for the ball to hot the ground. The ball hits the ground at when s(t) = 0. The equaton becomes:

0 = -4.9t^2 + 18.4t + 36.8

Multiply through by -10

49t^2 - 184t - 368 = 0

factorize the resulting expression

t = 18±√184²-4(49)(-386)/2(49)

t = 18±√338564+75656/78

t = 18±√414220/78

t = 18±643.599/78

t = 661.559/78

t = 8.48secs

Hence the object reaches that ground after 8.48seconds