Question

URGENT! TIMED QUESTION, WILL GIVE BRAINLIEST!

Answer 1

Answer:

Step-by-step explanation:

Part a

When the ball hits the ground  it has covered the height of zero meters so we replace h(t) with zero

$h(t)=-5t^2+40t\\0=-5t^2+40t\\40t-5t^2=0\\5t(8-t)=0\\5t=0 \ \ \ , \ \ \ 8-t=0\\t=0 \ \ \ , \ \ \ t=8$

So it will take 8 seconds for the ball to hit the ground we another value

t = 0 seconds is when the ball was thrown and after 8 seconds the balls hits the ground

Part c

The ball reaches its maximum height when the velocity is zero.

So as this functions represents the displacement of the ball h(t) we differentiate it to get the velocity equation, so we differentiate h(t) with respect to t.

$v=-10t+40$

so now we replace the velocity which is v with a zero because the ball reaches its maximum height when the velocity is zero.

$0=-10t+40\\10t=40\\t=40/10\\t=4$

so now we insert the value of t = 4 into the equation of h(t) to get the height.

$h(t)=-5t^2+40t\\h(4)=-5(4)^2+40(4)\\h(4)=-80+160\\h(4)=80$

so the maximum height of the ball is 80 meters.

Part d

The y-intercept of the graph is zero, and it represents that there is no initial height covered by the ball or the ball is actually thrown from already a height of x meters, as we can see from the equation the y-intercept is zero which means the height covered by the ball is 0 meters when time is zero.

So the point is actually at the Origin (0 , 0).

Part e

We can see that the origin (0 ,0) and (8,0) are the x-intercepts of the graph as  solved from part A

We can see that the stationary point or the maximum point is at (4,80) which means the maximum height of the ball is 80 meters when the time is 4 seconds as solved in part C

Graph attached.