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Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) A balloon, rising vert

Liula [17]

Answer:

(a) 2.79 seconds after its release the bag will strike the ground.

(b) At a velocity of 73.28 ft/second it will hit the ground.

Step-by-step explanation:

We are given that a balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 80 feet above the ground.

Assume the acceleration of the object is a(t) = −32 feet per second.

(a) For finding the time it will take the bag to strike the ground after its release, we will use the following formula;

$s=ut+\frac{1}{2} at^{2}$

Here, s = distance of the balloon above the ground = - 80 feet

u = intital velocity = 16 feet per second

a = acceleration of the object = -32 feet per second

t = required time

So, $s=ut+\frac{1}{2} at^{2}$

$-80=(16\times t)+(\frac{1}{2} \times -32 \times t^{2})$

$-80=16t-16 t^{2}$

$16 t^{2} -16t -80 =0$

$t^{2} -t -5 =0$

Now, we will use the quadratic D formula for finding the value of t, i.e;

$t = \frac{-b\pm \sqrt{D } }{2a}$

Here, a = 1, b = -1, and c = -5

Also, D = $b^{2} -4ac$ = $(-1)^{2} -(4 \times 1 \times -5)$ = 21

So, $t = \frac{-(-1)\pm \sqrt{21 } }{2(1)}$

$t = \frac{1\pm \sqrt{21 } }{2}$

We will neglect the negative value of t as time can't be negative, so;

$t = \frac{1+ \sqrt{21 } }{2}$ = 2.79 ≈ 3 seconds.

Hence, after 3 seconds of its release, the bag will strike the ground.

(b) For finding the velocity at which it hit the ground, we will use the formula;

$v=u+at$

Here, v = final velocity

So, $v=16+(-32 \times 2.79)$

v = 16 - 89.28 = -73.28 feet per second.

Hence, the bag will hit the ground at a velocity of -73.28 ft/second.

8 0

3 years ago

A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a p

irina [24]

Answer:

The boat is approaching the dock at rate of 2.14 ft/s

Step-by-step explanation:

The situation given in the question can be modeled as a triangle, please refer to the attached diagram.

A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow that means x = 5 ft.

The length of rope from bow to pulley is 13 feet that means y = 13 ft.

We know that Pythagorean theorem is given by

$x^{2} + y^{2} = z^{2}$

Differentiating the above equation with respect to time yields,

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$

$x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$

dx/dt = 0 since dock height doesn't change

$y\frac{dy}{dt} = z\frac{dz}{dt}$

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

The rope is being pulled in at a rate of 2 feet per second that is dz/dt = 2 ft/s

First we need to find z

z² = (5)² + (13)²

z² = 194

z = √194

z = 13.93 ft

So,

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

$\frac{dy}{dt} = \frac{13.93}{13}(2)$

$\frac{dy}{dt} = 2.14$ $ft/s$

Therefore, the boat is approaching the dock at rate of 2.14 ft/s

8 0

3 years ago

Solve for w in P = 2 w + 2 ℓ , if P = 38 and ℓ = 12 . <br><br> w = _____

olga nikolaevna [1]

Answer:

i think the answer would be 7 8/9

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The area of a circle is 50.24 square inches. What is the circle's radius?

hammer [34]

Area of a circle = pi* r^2
50.24 = pi* r^2
divide both sides by pi and get...
15.99188868 = r^2
the square root of that is 3.998985957 which rounds to 4

4 0

3 years ago

identify the maximum and minimum values of the function y=10cosx in the interval [-2pie, 2pie]. Use your understanding of tran

34kurt

Answer:

3 x + 2 y + z/ x + y + z , x = 2 , y = 3 , z = 1

tan ( x ) , x = − π

cot ( 3 x ) , x = 2 π /3

Step-by-step explanation:

3 0

3 years ago