Mathematics, 19.01.2021 14:00 EmanuelMarcos1572

On the planet Jupiter, a ball this thrown upward at a velocity at 10m/s. It's height, h(t) meters after t seconds is given by the equation h(t)=-12.5(t-0.4)^2+4. a) what is the maximum height the ball reaches?

b) what was the height of the ball when it was released from the throwers hand?

c) how many seconds after the ball was hit and did it land on the surface of Jupiter to the nearest second?

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Answer from: heart80941

a) The maximum height that the ball reaches is 4 meters.

b) The height of the ball when it is released from thrower's hand is 2 meters.

c) The ball will take approximately 0.966 seconds to hit the surface of Jupiter.

Step-by-step explanation:

a) Let $h(t) = -12.5\cdot (t-0.4)^{2}+4$ , where and h(t) are the time and height, measured in seconds and meters. Since this equation is in vertex form, the maximum height corresponds to the value associated with the dependent variable (height). That is:

$h(t) -4 = -12.5\cdot (t-0.4)^{2}$

Hence, the maximum height that the ball reaches is 4 meters.

b) The height of the ball when the ball is released from thrower's hand is the height of the ball at $t = 0\,s$ . Then, we evaluate the function:

$h(0) = -12.5\cdot (0-0.4)^{2}+4$

$h(0) = 2\,m$

The height of the ball when it is released from thrower's hand is 2 meters.

c) The instants when the ball hits the ground are those instants so that h(t) = 0 . Then:

$-12.5\cdot (t-0.4)^{2}+4 = 0$

$-12.5\cdot (t^{2}-0.8\cdot t +0.16)+4 = 0$

$-12.5\cdot t^{2}+10\cdot t +2 = 0$

By the Quadratic Formula we obtain the following roots:

$t_{1} \approx 0.966\,s$ , $t_{2}\approx -0.166\,s$

Only the first root is physically reasonable, since time is a positive real variable.

The ball will take approximately 0.966 seconds to hit the surface of Jupiter.

Answer from: Quest

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol

Answer from: Quest

answer:

total cost for tiles and paints is $924.

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

$\text{area of walls and ceiling}=(2*40*15)+(2*3*15)+(40*3)$

$\text{area of walls and ceiling}=1200+90+120$

$\text{area of walls and ceiling}=1410$

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

$\text{ the total painting cost}=10*(\frac{1410}{25})$

$\text{ the total painting cost}=10*56.4=564$

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

$\text{area of floor} = 40*3\text{ square meters}$

$\text{area of floor} =120\text{ square meters}$

given: 1m squared floor tiles costs $3. so,

$\text{total cost for tiles} = 3*120 = 360$

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

$\text{combined total cost}= 564+360 = 924$

therefore, the combined total cost of tiles and paint is $924.

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On the planet Jupiter, a ball this thrown upward at a velocity at 10m/s. It's height, h(t) meters af...

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