Question

1. A ball is dropped for the upper observation deck of a tower 450 meters above the ground. Its position above the ground at time t seconds since it was dropped is given by the funtion ?(?) =−4.9?² + 450.

A. What is the average velocity of the ball over the first three seconds?

B. What is the velocity of the ball 5 seconds after it was dropped?

c. How fast was the ball traveling when it hits the ground? (Hint: First, find the time that the ball hits the ground.)

d.The acceleration of the ball is constant at all times. What is that constant and where does

it come from?

2. The graph below shows the quadratic function ?(?) = −0.25?² + 3? + 5 and a tangent line to the function at ? = −2. Find the value of C, which is the x-intercept of the tangent line.

(Hint: First, find the equation of the line.)

С

Answer 1

(1)

we are given

s(t) = -4.9t2 + 450

(A)

we can use formula

$vavg=\frac{s(3)-s(0)}{3-0}$

now, we can plug values

vavg -4.9(3)2 + 450 - (-4.9(0)2 + 450) 3 - 0

........Answer

LVI- = bapa

(B)

s(t) = -4.9t2 + 450

we can find derivative

s(t) = -4.92t) + 0

now, we can plug t=5

$v(5)=-4.9(2(5))+0$

$v(5)=-49$..........Answer

(C)

we can set

s(t)=0

and then solve for t

$s(t)=-4.9t^2+450=0$

$t=9.583$

we can find derivative

s(t) = -4.92t) + 0

now, we can plug t=9.583

$v(9.583)=-4.9(2(9.583))+0$

$v(9.583)=-93.9134$..........Answer

(D)

s(t) = -4.92t) + 0

we can find derivative again to get accleration

$s''(t)=-4.9(2(1))+0$

$a(t)=-9.8$

It is the acceleration due to gravity

(2)

we are given

$f(x)=-0.25x^2+3x+5$

we can find derivative

$f'(x)=-0.25(2x)+3(1)+0$

now, we can plug x=-2

$f'(-2)=-0.25(2(-2))+3(1)+0$

$m=4$

now, we can find point

$f(-2)=-0.25(-2)^2+3(-2)+5$

$f(-2)=-2$

we can use point slope form of line

$y+2=4(x+2)$

we can solve for y

$y=4x+6$

we can set y=0

and then solve for x

$y=4x+6=0$

$x=-\frac{3}{2}$

$C=-\frac{3}{2}$............Answer