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2 years ago

HELP PLEASE LOOK AT PICTURE

Mathematics

1 answer:

diamong [38]2 years ago

6 0

As all lines are parallel to each other and HI is perpendicular to AB and LM is perpendicular to AB,
hence,
that angles are equal

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The equation y 4.9t 2 3.5t 2.4 relates the height y in meters to the elapsed time t in seconds for a ball thrown downward at 3.5

tamaranim1 [39]

Answer:

$t \approx 0.43\,s$

Step-by-step explanation:

The vertical displacement function is $y(t) = -4.9\cdot t^{2}-3.5\cdot t + 2.4$ , where $y(t)$ is measured in meters and $t$ in seconds. Ball hits the ground when $y(t) = 0$ . That is:

$-4.9\cdot t^{2}-3.5\cdot t + 2.4 = 0$

Whose roots can be found by using the General Formula for Second-Order Polynomials:

$t_{1,2} = \frac{3.5\pm \sqrt{(-3.5)^{2}-4\cdot (-4.9)\cdot (2.4)} }{2\cdot (-4.9)}$

Solutions of this polynomial are:

$t_{1} \approx 0.43\,s,t_{2} \approx -1.14\,s$

Only the first root is physically consistent.

3 0

3 years ago

What division problem does the model show?

garik1379 [7]

Answer:

54/3

Step-by-step explanation:

3 0

3 years ago

Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

$a_y$ = g = Acceleration due to gravity = $32.2\ \text{ft/s}^2$

t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$