Draw a diagram to illustrate the problem as shown below.

The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s

Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s

The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m

Answer: The height is 21.7 m (nearest tenth)

4 0

3 years ago

In the pair of supply and demand equations below, where x represents the quantity demanded in units of a thousand and p the unit

arlik [135]

Answer:

Equilibrium quantity = 5

Equilibrium price = 40

Explanation:

given:

p = -x²-3x+80

p = 7x+5

For the equilibrium quantity the price from both the functions will be equal

thus, we have

-x² - 3x + 80 = 7x+5

⇒ x² +3x + 7x + 5 - 80 = 0

⇒x² + 10x - 75 = 0

now solving for x

x²- 5x + 15x -75 = 0

x(x-5) + 15(x-5) = 0

therefore, the two roots of the equation are

x = 5 and x = -15

since the quantity cannot be in negative

therefore, the equilibrium quantity will be = 5

now the equilibrium price can be found out by substituting the equilibrium quantity in any of the equation

thus,

p = -(5)² -3(5) + 80 = 40

p = 7(5) + 5 = 40

4 0

2 years ago

Question 6

sleet_krkn [62]

Answer:

B: Process #1: Energy is decreasing Process#2: Energy is increasing

6 0

3 years ago

What is the kinetic energy of a 1500 kg object moving at a velocity of 10 m/s?

Anastasy [175]

The Kinetic Energy Formula is as follows: KE = mass x velocity^2 /2

Plug in the correct numbers into the variables!

KE = 1500 kg x 10 m/s ^2 / 2

Square the 10 m/s!

KE = 1500 kg x 100 m/s / 2

Multiply!

KE = 150,000 / 2

Divide!

KE = 75,000 Newton-meters or Joules!

4 0

3 years ago

Block and tackle is used only in crain why​

Block and tackle is used only in crain why