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Aleksandr [31]
3 years ago
11

A projectile is launched from ground level with an initial velocity of 89 ft/sec. It's height(ft), t seconds after launching is

given by s=-16t^2 + v0t + s0. a) Fine the time(s) the projectile will reach a height of 100 feet. b) find the time it takes the projectile to hit the ground
Mathematics
1 answer:
Otrada [13]3 years ago
7 0

Answer:

Step-by-step explanation:

Initial Velocity of projection = 89ft/sec

height(ft), t seconds after launching is given by s=-16t^2 + v0t + s0

A) time the projectile will reach a height of 100 ft

s=-16t^2 + v0t

v0 = 89ft/sec

s = 100ft

100 = - 16t² + 89t

-16t² + 89t - 100 = 0

Using the online quadratic equation solver;

t = 4s or t = 1.5625

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Step-by-step explanation:

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4 years ago
There is a beaker of 3.5% acid solution and a beaker of 6% acid solution in the science lab. Mr. Larson needs 200 milliliters of
oksian1 [2.3K]
Et x = the volume of the 3.5% solution 
<span>Let y = the volume 6% solution </span>

<span>(3.5%)x + (6%)y = (4.5%)(200) </span>
<span>0.035)x + 0.06y = 0.045(200) </span>
<span>0.035x + 0.06y = 9 ( Equation 1 )</span>

<span>x + y = 200 ( Equation 2 )</span>
<span>x = 200 - y </span>

<span>Substitute x = 200 - y into equation 1: </span>
<span>0.035x + 0.06y = 9 </span>
<span>0.035(200 - y) + 0.06y = 9 </span>
<span>7 - 0.035y + 0.06y = 9 </span>
<span>0.06y - 0.035y = 9 - 7 </span>
<span>0.025y = 2 </span>
<span>y = 2/(0.025) </span>
<span>y = 80 mL (the volume of 6% solution.)</span>

<span>Substitute y = 80 into equation 2: </span>
<span>x + y = 200 </span>
<span>x + 80 = 200 </span>
<span>x = 200 - 80 </span>
<span>x = 120 mL (the volume of the 3.5% solution.)</span>

<span>Therefore, Mr. Larson should combine 120 mL of the 3.5% solution </span>
<span>and 80 mL of the 6% solution to make 200 mL of 4.5% solution. </span>

<span>Answer is 120 mL of the 3.5% solution & 80 mL of the 6% solution

Hope I helped :)</span>
5 0
4 years ago
16-(2+6)•-8 solve with the order of operations
shusha [124]
The answer is 80 hope this helps :)

7 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!!!
Harlamova29_29 [7]

Answer:

D. y=\frac{1}{4}x

Step-by-step explanation:

3 0
3 years ago
A surveying team set up some equipment 104104 ft from the base of a tree in order to sight the top of the tree. From ground leve
jarptica [38.1K]

Answer:

Relative error = 0.0254 %

Step-by-step explanation:

Given details

Height of equipment from base is 104 ft

Elevation angle is 27.5 degree

From figure we have

tan\theta = \frac{x}{104} .........1

tan(27.5) = \frac{x}{104}

x  = 54.13 ft

differentiating 1st eq wrt x

\frac{d}{dx} tan\theta = \frac{d}{dx} [\frac{x}{104}]

sec^2\theta \frac{d\theta}{dx} = \frac{1}{104}

Taken d\theta \ as \Delta x \ and\  dx = \Delta x for minute change

\Delta x = 1.5% x

             = 0.015 \times 54.13 = 0.8119 ft

from sec^2 \theta \frac{d\theta}{dx} = \frac{1}{104}

     \Delta \theta = \frac{1}{104 sec^2 \theta} \Delta x

\Delta \theta = \frac{1}{104 \times 1.115} 0.8119

\Delta \theta  = 0.007

Reltaive error= \frac{\Delta \theta}{\theta} \times 100

                      = \frac{0.007}{27.5}\times 100

Relative error = 0.0254 %

4 0
4 years ago
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