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4 years ago

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An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in

seconds, the arrow is at a height of 500ft. Round your answer(s) to the nearest tenth of a second.

2 answers:

shtirl [24]4 years ago

7 0

Answer:

2.4 sec and 13.3 sec

Explanation:

SIZIF [17.4K]4 years ago

6 0

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

$h=-16t^2+v_ot$

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

$-16t^2+250t=500$

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

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A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f

zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

$v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s$

$v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s$

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

$s = v_vt + gt^2/2$

$10 = 19.97t - 9.8t^2/2$

$4.9t^2 - 19.97t + 10 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}$

$t= \frac{19.9658877511823\pm14.24}{9.8}$

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

$s_1 = v_ht_1 = 15*0.58 = 8.8 m$

$s_2 = v_ht_2 = 15*3.49 = 52.5m$

8 0

3 years ago

Which of the following statements about galaxies is true? A. Galaxies typically contain millions of stars. B. Galaxies actually

Sloan [31]

the answer is D i took the quiz so it is definitely D

6 0

3 years ago

Read 2 more answers

Do houses use parallels or series circuits

Vilka [71]

Every electrical outlet in your house, and every device or appliance that's
plugged into an outlet, are all in parallel. It's also most likely that all of yours
are in parallel with all the outlets, devices, and appliances in the homes or
apartments of a few of your neighbors.

The only things in your home that are connected in series are the switches
that turn things on and off.

3 0

3 years ago

Read 2 more answers

a large truck has a mass of 20000kg. it is traveling at 28m/s along a staright road . calculate the kinetic energy

Semmy [17]

Answer:

The answer is

<h2>7,840,000 J</h2>

Explanation:

The kinetic energy KE of an object given it's mass and velocity can be found by using the formula

$KE = \frac{1}{2} m {v}^{2} \\$

where

m is the mass

v is the velocity

From the question

m = 20000kg

v = 28 m/s

It's kinetic energy is

$KE = \frac{1}{2} \times 20000 \times {28}^{2} \\ = 10000 \times 784$

We have the final answer as

<h3>7,840,000 J</h3>

Hope this helps you

3 0

3 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago