Velocity stone question with specific numbers

Question

the height, in feet, of a stone thrown vertically upward from 44ft above ground level with an initial velocity of 40ft/s is h=-16t^2+40t+44. What is the velocity of the stone when it is 20ft above the ground?

Answer 1

velocity is the first derivative of the distance formula: v=-32t+40 max height is when t=40/32=5/4, h=69, v=0

20=-16t²+40t+44 divide by 4

5=-4t²+10t+11

-4t² + 10t +6 =0

-2t²+5t+3=0 solve with the quadratic formula

t=(-5 + or - (25-4(-2)(3))^1/2)/2(-2) = 5/4 + or - (7)/-4

=5/4 + or - -7/4 = -1/2 or 3 this is the time when height is 20, ignore the negative t

Now substitute t=3 into the velocity equation v=-32t+40

v=-32(3)+40 = -96+40=-56 feet per second, a negative speed indicating going down

-56 is the speed going down after first going up and reaching at maximum height where velocity =0 , then falling

It starts at height 44, reaches maximum height 69 and falls to height 20 at times 0, 5/4, and 3 seconds

at 2 1/2 seconds or 5/2 seconds, the height is the same as at the beginning, 44 feet. In the next 1/2 second it falls to 20 feet. You can graph the motion up and down as a parabola, symmetric around x=5/4 opening downward, with vertex at the maximum height 69 = -16(5/4)²+40(5/4)+44 = -25+50+44= 69

Answer 2

V = Vo +g*t

V = velocity

Vo- initial velocity

g- gravity constant

t- time

Given Vo= 40ft/s

Hf = height final = 44ft ( tallest point)

Hx = 20 ft or wanted height

H(t) = velocity in terms of time is given

h=-16t^2+40t+44

Set height to 20 ft

20=-16t^2+40t+44 solve for t

t = -1/2 and 3

We cannot have negative time so we can ignore that solution and just use 3

H= V^2 - (Vo / (2*g)) (equation from physics law)

Rearrange for V

H + (Vo / 2*g) = V^2

V = sqrt ( H + (Vo / 2*g) )

V= sqrt (20ft + ((44ft/s) / (2 *32.174505 ft/s^2)

V= 20.68 ft/s