The number of seconds it takes until the ball reaches max height is 2 seconds and the maximum height is 112 feet, and the number of seconds it takes until the ball hits the ground will be 6.9 seconds.
When a body is thrown in the air then the motion under the gravity is known as projectile motion.
A person standing close the the edge on the top of a 80 foot building throws a baseball vertically upward. the quadratic function
h(t) = -8t² + 32t + 80 models the ball's height above the ground, h(t), in feet, t seconds after it was thrown.
The number of second to reach at maximum height will be
We know that at maximum height, the rate will be zero. Then we have
h'(t) = 0
-16t + 32 = 0
t = 2 seconds
Then the maximum height will be
h(t) = -8(2)² + 32 x 2 + 80
h(t) = -32 + 64 +80
h(t) = -32 + 144
h(t) = 112 feet
Then the number of the seconds to reach at the ground will be
-8t² + 32t + 80 = -80
-8t² + 32t + 160 =0
t² - 4t - 20 = 0
Then t will be
[tex]\rm t = \dfrac{-(4) \pm \sqrt{(-4)^2 - 4 * 1 * -160}}{2*1}[/tex]
Then we have
t = 6.9 and -2.9
More about the projectile motion link is given below.
https://brainly.com/question/11049671
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