What's my name A. Lou Sahsshole B. Dihcksl Ong C. Gabe Itch D. Moe Lester

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

3 years ago

What is the sum of this geometric series?24 + 36 + ... + 121.5

krok68 [10]

The sum of the geometric series is 195

4 0

4 years ago

I now have another question to ask.

marissa [1.9K]

Answer:

23.6 ft

Step-by-step explanation:

Sketch a right triangle representing this situation. The length of the hypotenuse is 26 ft and the angle of elevation from ground to top of ladder is 65°. The "opposite side" is the reach of the ladder, which we'll call x.

Then:

opp

sin 65° = ----------
26 ft

or (26 ft)(sin 65°) = opp side = height off the ground of top of ladder.

Evaluating this, we get:

(26 ft)(0.906) = 23.56 ft, or, rounded off, 23.6 ft

The ladder reaches 23.6 ft up the side of the building.

8 0

3 years ago

What are the common factors of 24 64 88

8090 [49]

Answer is 8

8 times 3 is 24

8 times 8 is 64

8 times 11 is 88

7 0

3 years ago

a port and a radar station are 2 mi apart on a straight shore running east and west. a ship leaves the port at noon traveling no

Rom4ik [11]

Answer:

The rate of change of the tracking angle is 0.05599 rad/sec

Step-by-step explanation:

Here the ship is traveling at 15 mi/hr north east and

Port to Radar station = 2 miles

Distance traveled by the ship in 30 minutes = 0.5 * 15 = 7.5 miles

Therefore the ship, port and radar makes a triangle with sides

2, 7.5 and x

The value of x is gotten from cosine rule as follows

x² = 2² + 7.5² - 2*2*7.5*cos(45) = 39.04

x = 6.25 miles

By sine rule we have

$\frac{sin A}{a} = \frac{sin B}{b}$

Therefore,

$\frac{sin 45}{6.25} = \frac{sin \alpha }{7.5}$

α = Angle between radar and ship α

∴ α = 58.052

Where we put

$\frac{sin 45}{6.25} = \frac{sin \alpha }{x}$ to get

$\frac{x}{6.25} = \frac{sin \alpha }{sin 45}$ and differentiate to get

$\frac{\frac{dx}{dt} }{6.25} = cos\alpha\frac{\frac{d\alpha }{dt} }{sin 45}$

$\frac{15sin45 }{6.25cos\alpha } =\frac{d\alpha }{dt} }$ = 3.208 degrees/second = 0.05599 rad/sec.

6 0

3 years ago