## Alberta Free Tutoring And Homework Help For Math 20-2

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### Bill and Ben are on a bridge, timing how long it takes stones they have dropped to hit the water below.  A quadratic relation can be used to determine the distance, D, in metres, that ta stone will fall in the time, t, in seconds after it is realeased.  In this relation, a = -0.5g, where g is acceleration due to gravity, which is approximately 9.8 m/s2 on Earth.  Ben starts a timer when Bill releases a stone, and he stops the timer when the stone hits the water below.  The mean time of serveral trials is 3 s.a) Determine a quadratic function that models the falling stones.b) How high above the water are Bill and Ben? Explain your answer.c) On Saturn's moon, Titan, the value of g is 1.35 m/s2.  Suppose that Bill and Ben are astronauts and they are standing at the top of a cliff on Titan.  If they record the time for a stone to fall from their hands to the bottom of the cliff as 3 s, how high is the cliff, to the nearest hundreth of a metre?

3 years ago

#### Answered By Geoff B

a)  y = a * x^2

y = (-0.5 * 9.8(m/s^2) * x^2

y = -4.9 (m/s^2) * x^2

y = height (m),        x=time (s)

b)  y = -4.9 (m/s^2) * (3 s)^2

=  -44 m

the height of the bridge is 44m because the ball would fall down 44 m in 3 s according to the quadratic equation

c)   y = (-0.5 * 1.35 (m/s^2) * x^2

= -0.675 (m/s^2) * (3 s)^2

= -6.1 m

cliff is 6.1 m tall

3 years ago

#### Answered By Michael Z

a) so this part of the question you need to determine the relation of distance (d) to time (t) and acceleration(a)

knowing that this is a quadratic equation we know that there has to be a squared in there somewhere. basically we need to figure out which variables (d, t, or a ) are dependent and independent. The independent variable will be the term with the 2 on it. If we look in the question, variable a is given which is -0.5xg and g is given as 9.8m/s2. This leaves us with d or t. We know that time (t) will always be independent which leaves us with d as dependent. So putting everything in the equation together gives us

d=at2

d=-0.5gt2

d=-0.5x9.8xt2

b)so here we need to use our equation from part a and figure out the height. We know that the average time given is 3s. Plug it in to the equation and solve. you should get 44.1m. The quadratic equation in part a models the motion of the ball when it is falling under the influence of gravity. It will fall 44.1m in 3 secs. The answer you will get will be a negative one. The negative sign is from physics which tells you the direction in which the ball is falling. In this case a negative sign means the ball is falling down.

c) repeat the logic used in part a and b but g will now be 1.35m/s2 instead of 9.8m/s2

you should get:

d=-0.5 x 1.35 x t2  for your equation and 6.08m for the height.

The negative sign that you would get for the height at -6.08m comes from physics and it tells you the direction in which the ball is falling. If its negative the ball is falling down. If its positive it would be going up. Since we ended up with a negative sign we know that we are right in having the ball fall down.

3 years ago

#### Answered By Jason R

1.

The general form of a quadratic equation is y = ax2 + bx + c

We are told that D(distance) depends on t(time) and a.

Since b and c are ommited we can assume the have a value of 0.

We're left with: D = at2 + 0t + 0

Simplified: D = at2

Written in function notation: D(t) = at2

2.

From the question we are told:

a = 0.5*9.8m/s2 = 4.9m/s2

t = 3s

t2= 3s*3s = 9s2

Putting it all together we have

D = at2= 4.9m/s2*9s2 = 44.1m

The s2(seconds squared) term cancles out leaving only the m(meter) unit left which is good because that is exactly what we're looking for.

The cliff is 44.1m high.

3.

From the question:

a = 0.5g = 0.5*1.35m/s2 =

t = 3s

t2= 3s*3s = 9s2

All that is left is the substitute the value into our quadratic equation.

D = at2 = 0.675m/s2*9s2 = 6.075m

We are asked to give the answer to the nearest hundreth. Notice again how the s2 cancel out and leave m.

D = 6.08m