Question

A softball is thrown into the air with an initial velocity of 5 meters per second from a height of 9 meters.

The function below models the distance of the softball from the ground in meters after t seconds.

How many seconds does it take for the softball to hit the ground?


h(t)=4.9t^2+5t+9

Answer 1

0.94secs

Explanation:

Given the function below that models the distance of the softball from the ground in meters after t seconds expressed as;

h(t)=-4.9t^2+5t+9

The ball hits the ground when the height h(t) is zero

Substitute h(t) = 0 into the expression and calculate the value of t;

0 = -4.9t²+5t+9

Multiply through by 10;

0 = 49t²+50t+90

-49t²+50t+90 = 0

Factorize;

t = -50 ±√50²-4(49)(90)/2(49)

t = -50±√2500+17640/98

t = -50±√20140/98

t = -50+141.92/98

t = 91.92/98

t = 0.94 secs

Hence it takes 0.94secs for the softball to hit the ground