Question

1

1 point
An object is launched at 12.5 meters per
second from a 24.8 meter tall platform.
The equation for the object's height is
h(t) = -3.21 + 12.5t + 24.8.
When does the object strike the ground?
-2
5.7
ООО
8
O


Answer 1

5.33 secs

Explanation:

Given the equation of the height modeled by the equation/

h(t) = -3.2 + 12.5t + 24.8

The object strikes the ground when the height h(t) is zero

Substituting h(t) = 0 into the expression

h(t) = -3.2t² + 12.5t + 24.8

0 = -3.2t² + 12.5t + 24.8

-3.2t² + 12.5t + 24.8 = 0

Multiply through by minus sign

3.2t² - 12.5t - 24.8 = 0

From the expression a = 3.2, b = -12.5 and c = -24.8

t = -(-12.5)±√(-12.5)²-4(3.2)(-24.8)/2(3.2)

t = 12.5±√156.25+317.44)/6.4

t = 12.5±√473.69/6.42

t = 12.5±21.76/6.42

t = 12.5+21.76/6.42

t = 34.26/6.42

t = 5.33secs

Hence the object strike the ground after 5.33 secs