Mathhelp346
  • Mathhelp346
PLEASEE help with attached: How do I complete the table shown on the second picture?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mathhelp346
  • Mathhelp346
Mathhelp346
  • Mathhelp346
Activity 1 is the first picture
anonymous
  • anonymous
what is the velocity of projection???

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Mathhelp346
  • Mathhelp346
1 Attachment
Mathhelp346
  • Mathhelp346
Does that help?
anonymous
  • anonymous
yep
anonymous
  • anonymous
do u know the equations of motion??
Mathhelp346
  • Mathhelp346
What do you mean?
anonymous
  • anonymous
okay leave it
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
time taken to reach the maximum height t = u/g
anonymous
  • anonymous
on earth surface , t = 24/32 =3/4 second
Mathhelp346
  • Mathhelp346
So it's like \[v=\frac{ d }{ t }\]Where d is distance and t is time?
anonymous
  • anonymous
no do u know the formula V = U + gt
Mathhelp346
  • Mathhelp346
No
anonymous
  • anonymous
u r of which standard??
Mathhelp346
  • Mathhelp346
?
anonymous
  • anonymous
class of study?
Mathhelp346
  • Mathhelp346
8th grade?
anonymous
  • anonymous
do u know about displacement,velocity,acceleration etc
Mathhelp346
  • Mathhelp346
Yes
anonymous
  • 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
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
now ur first question maximum height of the ball
anonymous
  • anonymous
at maximum heicht, final velocity is zero
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
so, 0=U+at t = -U/a
anonymous
  • anonymous
here acceleration is nothing but acceleration due to gravity
anonymous
  • anonymous
when a body is projected vertically upwards acceleation due to gravity is taken negetive
anonymous
  • anonymous
so finally t =U/g
Mathhelp346
  • Mathhelp346
What do you mean acceleration is nothing?
anonymous
  • anonymous
acceleration is equal to acceleration due to gravity in this case
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
so finally time required to reach maximum height t = U/g = 24/32 =3/4 second
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
so u can do this on remaining planets
Mathhelp346
  • Mathhelp346
So Mercury would be 2 seconds right
anonymous
  • anonymous
2.02
anonymous
  • anonymous
next maximum height reached by the ball
Mathhelp346
  • Mathhelp346
.85 seconds
anonymous
  • anonymous
for what?
Mathhelp346
  • Mathhelp346
Venus
anonymous
  • anonymous
correct
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
do the remaining later lt's go for the next question
Mathhelp346
  • Mathhelp346
Ok
Mathhelp346
  • Mathhelp346
Wait how do you find the maximum height?
anonymous
  • anonymous
for maximum height reached substitute t = U/g in h(t)
Mathhelp346
  • Mathhelp346
You said it was t = -U/a?
Mathhelp346
  • Mathhelp346
So I substitute the time required to reach maximum height into the quadratic functions?
anonymous
  • anonymous
ya
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
u will get h(t) = U^2/2g + ho
Mathhelp346
  • Mathhelp346
Ok
Mathhelp346
  • Mathhelp346
So the answer to the first one would be 27 ft right
anonymous
  • anonymous
now on earth surface, h = (24)^2/2(32) +3 = 12
anonymous
  • anonymous
ya right 27.32
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
now u can do the remaining let's move to the next question
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
time required to return to planet's surface
anonymous
  • 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
Mathhelp346
  • Mathhelp346
Ok
Mathhelp346
  • Mathhelp346
So it's the same answer?
anonymous
  • anonymous
no
anonymous
  • anonymous
wait this is some what lengthy
anonymous
  • anonymous
now, time taken by the ball to reach the hand of the person is, U/g + U/g = 2U/g
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
but the ball should reach the ground
anonymous
  • 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
anonymous
  • anonymous
so it reaches the hands of the person with a velocity of 24 feet per second
anonymous
  • anonymous
now it should reach the ground by traveling 3 feet
Mathhelp346
  • Mathhelp346
So 27 feet
anonymous
  • anonymous
no how do u say that?
anonymous
  • anonymous
u der?
Mathhelp346
  • Mathhelp346
Yes
anonymous
  • anonymous
now h(t) = (1/2)g(t^2) + Ut 3 = 16t^2 + 24t
anonymous
  • anonymous
t = 0.116 seconds
anonymous
  • anonymous
now add this with 2U/g
anonymous
  • anonymous
hope u understood
Mathhelp346
  • Mathhelp346
There's a negative in front of (1/2)g And how did you get 3?
anonymous
  • anonymous
that is given because there the ball is projected vertically upward here the ball is coming down
anonymous
  • anonymous
it should travel 3 feet downward to reach the ground
Mathhelp346
  • 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?
anonymous
  • anonymous
ya
Mathhelp346
  • Mathhelp346
So does that mean they all travel 3 feet downward?
anonymous
  • anonymous
no
Mathhelp346
  • Mathhelp346
but the ho for each of them is 3
anonymous
  • anonymous
the ball should travel 3 feet downward after it reaches the hands of the person ,to reach the ground
Mathhelp346
  • Mathhelp346
So how long do the other ones travel?
anonymous
  • anonymous
same 3 feet
Mathhelp346
  • Mathhelp346
So the other ones travel 3 feet downward?
anonymous
  • anonymous
|dw:1355813142620:dw|
Mathhelp346
  • Mathhelp346
Ok
anonymous
  • anonymous
i think now u can understand clearly
anonymous
  • anonymous
okay?
Mathhelp346
  • Mathhelp346
Ok thank you
anonymous
  • anonymous
welcome
anonymous
  • anonymous
:)
Mathhelp346
  • Mathhelp346
Mercury's time would be 4 seconds?

Looking for something else?

Not the answer you are looking for? Search for more explanations.