Question

The equation $y = -4.9t^2 + 42t + 18.9$ describes the height (in meters) of a ball tossed up in the air at 42 meters per second from a height of 18.9 meters from the ground, as a function of time in seconds. In how many seconds will the ball hit the ground?

Answer 1

9 seconds

Explanation:

y = -4.9t² +42t +18.9

if the ball will be on the ground then y =0 and we need to find t by solving the quadratic equation

-4.9t² +42t +18.9 = 0

t= (-b±√b²-4ac)/2a, quadratic formula

t= {-42±√[42²-4*(-4.9)(18.9)] }/ 2(-4.9) , use calculator to solve

t= (-42 ±√2134.44)/ -9.8, solve the square root

t= (-42 ±46.2)/-9.8

one t = (-42 - 46.2)/-9.8 = - 88.2/ -9.8 = 9 seconds

another t = (-42 + 46.2)/-9.8 = 4.2 / -9.8 ≈ - 0.43 seconds reject this answer