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Mathematics 20 Online
OpenStudy (anonymous):

Jules kicks a soccer ball off the ground and into the air with an initial velocity of 25 feet per second. Assume the starting height of the ball is 0 feet. Approximately what maximum height does the soccer ball reach? 0.8 ft 1.6 ft 9.8 ft 19.6 ft Approximately, how long does it take until the soccer ball hits the ground again? 0.6 sec 0.8 sec 1.6 secs 2.8 secs

OpenStudy (anonymous):

Assuming that Jules kicks the ball completely vertically, then you can use one of the famous kinematics equations (I'll spare you the derivation here): (v-final)^2 - (v-initial)^2 = 2*a*height In your case, the ball's final velocity (at its maximum height) is zero! Your acceleration is -g = -32 feet per second per second, and your initial velocity is 25 feet per second. So you have: height = - (25)^2 / (2*-32) = 625 / 64 = 9.8 feet

OpenStudy (cwrw238):

at maximum height velocity v = u = initial velocity = 25, s = height, a = -32 ft/s/s v^2 =u ^2 + 2 as 0 = 25^2 + 2 * -32 * s 0 = 625 - 16s 16s = 625

OpenStudy (cwrw238):

aawwww - i mucked that up its 0 = 25^2 - 64s s = 625/64

OpenStudy (cwrw238):

yes its 9.8 ft

OpenStudy (anonymous):

So he was right. The answer to the first one is 9.8ft. But what about the second question?

OpenStudy (cwrw238):

use another constant acceleration equation s = displacement = 0, u = 25, a = -32, t = ? s = ut + 0.5at^2 0 = 25t - 16t^2 16t^2 - 25t = 0 t(16t - 25) = 0 can you solve this megan?

OpenStudy (anonymous):

Is it 1.6?

OpenStudy (cwrw238):

yes!! good one

OpenStudy (cwrw238):

t = 25/16 = 1.6 to 1 D P

OpenStudy (anonymous):

Can you help me with two more of these problems? i would really appreciate it!

OpenStudy (cwrw238):

ok

OpenStudy (anonymous):

Amir pitched a baseball at an initial height of 5.9 feet with a velocity of 73.16 feet per second. About how long will it take the ball Amir pitched to hit the ground? 4.66 secs 2.94 secs 2.35 secs 1.47 secs

OpenStudy (anonymous):

S = Vit + ½at² 5.9 = 73.16t + ½(32.2)t² 16.2t² + 73.16 t - 5.9 = 0 t = [-(73.16) ±√73.16² - 4(16.1)(-5.9)]/2(16.1) t = -73.16 ± 75.71 / 32.2 t = 0.079 sec & 4.6 sec

OpenStudy (cwrw238):

wow - thats fast!!

OpenStudy (anonymous):

Thanks!

OpenStudy (anonymous):

Ken is building a table. He attached planks of wood and now wants to build a border around the planks. The area of the entire table should be no larger than 3,276 square inches. All together, the planks are 36 inches wide and 72 inches long. Which quadratic equation represents the area of the planks and border combined? (The border must have equal width on each side.) 4x2 + 216x + 2,592 = 0 4x2 + 216x − 684 = 0 2x2 + 216x − 3,276 = 0 x2 + 108x + 3,276 = 0

OpenStudy (cwrw238):

i'm looking at deal or no deal - exciting bit - so if lawrence hasn't done it in record time i'll have a go

OpenStudy (anonymous):

I'm kind of in a rush....could you please help me

OpenStudy (anonymous):

@lawrenceh123 please help

OpenStudy (cwrw238):

om - im a bit confused wit with this one the area of one plank = 36 * 72 = 2592

OpenStudy (cwrw238):

so i guess the first one would be the answer

OpenStudy (anonymous):

4x2 + 216x + 2,592 = 0?

OpenStudy (anonymous):

If he makes the border x inches wide, then there will be an area of 36*x of border at each end and an area of 72*x of border on each longer side. At each of the 4 corners, there will be an additional x*x border area. So the entire area of a maximum-size tabletop is 4*x*x + 2*36x + 2*72x + 36*72 = 3,276 square inches. Simplifying this, we get 4x^2 + 216x + 2,592 = 3,276 4x^2 + 216x − 684 = 0 which is the second proposed answer.

OpenStudy (cwrw238):

ahhh - yes - good work lawrence

OpenStudy (anonymous):

ty

OpenStudy (anonymous):

ty

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