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2 years ago

A line goes through (2, 4) and has a slope of 1/2.

Mathematics

2 answers:

OverLord2011 [107]2 years ago

7 0

Answer:

y-y1=m(x-x1)

y-4=1/2(x-2)

y=1/2x+3

Harlamova29_29 [7]2 years ago

6 0

Answer:

Step-by-step explanation:

m = 1/2

Equation of line: y - y₁ = m(x - x₁)

x₁ = 2 & y₁ = 4

$y -4 = \dfrac{1}{2}(x - 2)\\\\y - 4 =\dfrac{1}{2}x -2*\dfrac{1}{2}\\\\y-4=\dfrac{1}{2}x -1\\y = \dfrac{1}{2}x -1+4\\\\y=\dfrac{1}{2}x+3$

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2 years ago

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mojhsa [17]

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(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

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Angle of projection is, $\theta=25$ °

Initial speed is, $u=15\ m/s$

Time of flight is, $t=3.0\ s$

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

$u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s$

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = $-h$ (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is, $g=-9.8\ m/s^2$

Now, using the following equation of motion;

$-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m$

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So, $v_y=0\ m/s$

Now, using the following equation of motion, we have:

$v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m$

Therefore, the maximum height of the brick is 2.05 m.

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