Question

The quadratic function h(t)=-16.1t^2 + 150 models a balls height, in feet, over time, in seconds, after it is dropped from a 15 story building.
- From what height, in feet, was the ball dropped?
-After how many seconds, rounded to the nearest hundredth, did the ball hit the ground?

Answer 1

In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05 

Answer 2

The time it takes the ball to hit the ground is 3.05secs

Calculating the height of the ball

Given the equation of the height that models the ball expressed as:

$h(t)=-16.1t^2 + 150$

At t = 0, the height is expressd as:

h(0) = -16.1(0)^2 + 150

h(0) = 150ft

Hence the height in feet of the ball is 150 feet

The height of the ball on the ground is 0ft. Hence;

0 = -16.1t^2 + 150

t^2 = 150/16.1

t^2 = 9.31

t = 3.05secs

Hence the time it takes the ball to hit the ground is 3.05secs

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