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Mathematics 14 Online
OpenStudy (mathhelp346):

PLEASEE help with attached: How do I complete the table shown on the second picture?

OpenStudy (mathhelp346):

OpenStudy (mathhelp346):

Activity 1 is the first picture

OpenStudy (anonymous):

what is the velocity of projection???

OpenStudy (mathhelp346):

OpenStudy (mathhelp346):

Does that help?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

do u know the equations of motion??

OpenStudy (mathhelp346):

What do you mean?

OpenStudy (anonymous):

okay leave it

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

time taken to reach the maximum height t = u/g

OpenStudy (anonymous):

on earth surface , t = 24/32 =3/4 second

OpenStudy (mathhelp346):

So it's like \[v=\frac{ d }{ t }\]Where d is distance and t is time?

OpenStudy (anonymous):

no do u know the formula V = U + gt

OpenStudy (mathhelp346):

No

OpenStudy (anonymous):

u r of which standard??

OpenStudy (mathhelp346):

?

OpenStudy (anonymous):

class of study?

OpenStudy (mathhelp346):

8th grade?

OpenStudy (anonymous):

do u know about displacement,velocity,acceleration etc

OpenStudy (mathhelp346):

Yes

OpenStudy (anonymous):

okay so now u learn this formula V= U + at V= final velocity of the body U= initial velocity of the body a = acceleration t = time

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

now ur first question maximum height of the ball

OpenStudy (anonymous):

at maximum heicht, final velocity is zero

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

so, 0=U+at t = -U/a

OpenStudy (anonymous):

here acceleration is nothing but acceleration due to gravity

OpenStudy (anonymous):

when a body is projected vertically upwards acceleation due to gravity is taken negetive

OpenStudy (anonymous):

so finally t =U/g

OpenStudy (mathhelp346):

What do you mean acceleration is nothing?

OpenStudy (anonymous):

acceleration is equal to acceleration due to gravity in this case

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

so finally time required to reach maximum height t = U/g = 24/32 =3/4 second

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

so u can do this on remaining planets

OpenStudy (mathhelp346):

So Mercury would be 2 seconds right

OpenStudy (anonymous):

2.02

OpenStudy (anonymous):

next maximum height reached by the ball

OpenStudy (mathhelp346):

.85 seconds

OpenStudy (anonymous):

for what?

OpenStudy (mathhelp346):

Venus

OpenStudy (anonymous):

correct

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

do the remaining later lt's go for the next question

OpenStudy (mathhelp346):

Ok

OpenStudy (mathhelp346):

Wait how do you find the maximum height?

OpenStudy (anonymous):

for maximum height reached substitute t = U/g in h(t)

OpenStudy (mathhelp346):

You said it was t = -U/a?

OpenStudy (mathhelp346):

So I substitute the time required to reach maximum height into the quadratic functions?

OpenStudy (anonymous):

ya

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

u will get h(t) = U^2/2g + ho

OpenStudy (mathhelp346):

Ok

OpenStudy (mathhelp346):

So the answer to the first one would be 27 ft right

OpenStudy (anonymous):

now on earth surface, h = (24)^2/2(32) +3 = 12

OpenStudy (anonymous):

ya right 27.32

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

now u can do the remaining let's move to the next question

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

time required to return to planet's surface

OpenStudy (anonymous):

if the ball is projected from the ground time taken to reach the maximum height = time taken to reach the ground from maximum height

OpenStudy (mathhelp346):

Ok

OpenStudy (mathhelp346):

So it's the same answer?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

wait this is some what lengthy

OpenStudy (anonymous):

now, time taken by the ball to reach the hand of the person is, U/g + U/g = 2U/g

OpenStudy (mathhelp346):

Ok

OpenStudy (anonymous):

but the ball should reach the ground

OpenStudy (anonymous):

there one more property that, if a ball is projected vertically up with certain velocity ,the ball acquire the same velocity when it reaches the original position

OpenStudy (anonymous):

so it reaches the hands of the person with a velocity of 24 feet per second

OpenStudy (anonymous):

now it should reach the ground by traveling 3 feet

OpenStudy (mathhelp346):

So 27 feet

OpenStudy (anonymous):

no how do u say that?

OpenStudy (anonymous):

u der?

OpenStudy (mathhelp346):

Yes

OpenStudy (anonymous):

now h(t) = (1/2)g(t^2) + Ut 3 = 16t^2 + 24t

OpenStudy (anonymous):

t = 0.116 seconds

OpenStudy (anonymous):

now add this with 2U/g

OpenStudy (anonymous):

hope u understood

OpenStudy (mathhelp346):

There's a negative in front of (1/2)g And how did you get 3?

OpenStudy (anonymous):

that is given because there the ball is projected vertically upward here the ball is coming down

OpenStudy (anonymous):

it should travel 3 feet downward to reach the ground

OpenStudy (mathhelp346):

\[h(t)=-\frac{ 1 }{ 2 }g t ^{2}+v _{0}t +h _{0}\] So you got the 3 from Earth's quadratic function right?

OpenStudy (anonymous):

ya

OpenStudy (mathhelp346):

So does that mean they all travel 3 feet downward?

OpenStudy (anonymous):

no

OpenStudy (mathhelp346):

but the ho for each of them is 3

OpenStudy (anonymous):

the ball should travel 3 feet downward after it reaches the hands of the person ,to reach the ground

OpenStudy (mathhelp346):

So how long do the other ones travel?

OpenStudy (anonymous):

same 3 feet

OpenStudy (mathhelp346):

So the other ones travel 3 feet downward?

OpenStudy (anonymous):

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