need help ASAP with algebra II/physics question

Question

Question: Eli stands on a rooftop of a 5 story building that is 50 feet above the ground, and throws a balloon straight up at a speed of 10 ft/sec In parts a-e, round your answers to the nearest thousandths.

a) write the equation for the height of the balloon in terms of time since Eli threw it

b) when does the balloon hit the ground?

c) when does the balloon pass the point from which it was released?

d) when does the balloon pass by a third-story window at 24 feet?

e) what is the maximum height reached by the ball? when does it reach this height?

Answer 1

For these problems, use the kinematic equation y = yo + vot - (1/2)gt²

where yo is the initial height of the ball, vo is the initial velocity, g = 32.2 ft/sec² is the acceleration due to gravity, and t is the time.

Assume that positive y is up and that the ground corresponds to y = 0. Then. yo = 50 feet and vo = +10 feet/sec. The last term is negative because gravity points toward the ground, or down.

So, the kinematic equation becomes

y = 50 + 10t -(1/2)(32.2)t²

(a) y = 50 + 10t - 16.1t²

(b) When is y = 0?

-16.1t² + 10t + 50 = 0

Solve this quadratic. Multiply through by -1.

16.1t² - 10t - 50 = 0

Use the quadratic formula.

t = {10 +/- [100 - 4(16.1)(-50)]^1/2} / [(2)(16.1)]

t = {10 +/- [100 + 3220]^1/2} / (32.2)

t = {10 +/- 57.62} / (32.2)

Discard the negative root since time must be positive.

t = {10 + 57.62} / (32.2)

t = 2.1 sec, the time that the ball strikes the ground

(c) When is y = 50, the initial height?

Go back to master equation in (a) and set y to 50.

y = 50 + 10t - 16.1t²

-16.1t² + 10t + 50 = 50

The 50's cancel.

-16.1t² + 10t = 0

Multiply by -1.

16.1t² - 10t = 0

t(16.1t - 10) = 0

t = 0 and 16.1t = 10, or t = 0.62 sec

The time t = 0 is the initial time. The later time t = 0.62 sec is when the ball passes by the initial height on its way toward the ground.

t = 0.62 sec

(d) When is y = 24? Set y = 24 in the master equation in from part (a).

-16.1t² + 10t + 50 = 24

-16.1t² + 10t + 26 = 0

Multiply through by -1.

16.1t² - 10t - 26 = 0

Solve using the quadratic formula.

t = {10 +/- [100 - 4(16.1)(-26)]^1/2} / [(2)(16.1)]

t = {10 +/- [100 + 1674.4]^1/2} / (32.2)

t = {10 +/- 42.12} / (32.2)

Discard the negative root since time must be positive.

t = {10 + 42,12} / (32.2)

t = 1.62 sec, the time that the ball is 24 feet above the ground

(e) In part (c) we saw that it took 0.62 sec for the ball to go up and come back to its original release point. Half of this time was spent rising. So, the maximum height was at t = 0.62 sec / 2 = 0.31 sec

At t = 0.31 sec, we can use the master equation from part (a) to find y.

y = 50 + 10t - 16.1t²

y = 50 + 10(0.31) - 16.1(0.31)²

y = 51.55 feet, maximum height reached.

Answer 2

To answer this question,we must first assume that Eli throws a ball, which is not influenced by air resistance. This will not work for a helium balloon.

a) the height of the ball is z = z0 + v0*t + 1/2 a *t ^2, where z0 is the initial height (50 feet), v0 is the initial vertical speed (10 ft/sec) and a is the acceleration due to gravity (-32 ft/sec^2)

b) ball hits the ground when z = 0; so 0 = 50 + 10t -16t^2. Reorganize to its more familiar form

-16t^2 + 10t + 50 = 0, and solve using the quadratic formula

c) essentially the same problem, but now z = 50; so 50 = 50 + 10t - 16t^2. This reduces to

16t=10 and t = 10/16 = 0.62 seconds.

d) same problem again, with z = 24, so 24 = 50 + 10t - 16 t^2. Reorganize to the familiar form

-16t^2 +10t + 26 = 0, and solve using the quadratic formula

e) The ball slows as it rises; when it reaches its maximum height, its vertical speed is zero. The velocity formula is v = v0 + at or v = 10 - 32t. When v = 0, t = 10/32 = 0.31 seconds. Note that this is half the time it takes for the ball to return to its initial height.