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Mathhelp346
Group Title
PLEASEE help with attached:
How do I complete the table shown on the second picture?
 one year ago
 one year ago
Mathhelp346 Group Title
PLEASEE help with attached: How do I complete the table shown on the second picture?
 one year ago
 one year ago

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Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
Activity 1 is the first picture
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
what is the velocity of projection???
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
Does that help?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
do u know the equations of motion??
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
What do you mean?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
okay leave it
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
time taken to reach the maximum height t = u/g
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
on earth surface , t = 24/32 =3/4 second
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
So it's like \[v=\frac{ d }{ t }\]Where d is distance and t is time?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
no do u know the formula V = U + gt
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
u r of which standard??
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
class of study?
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
8th grade?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
do u know about displacement,velocity,acceleration etc
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
now ur first question maximum height of the ball
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
at maximum heicht, final velocity is zero
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
so, 0=U+at t = U/a
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
here acceleration is nothing but acceleration due to gravity
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
when a body is projected vertically upwards acceleation due to gravity is taken negetive
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
so finally t =U/g
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
What do you mean acceleration is nothing?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
acceleration is equal to acceleration due to gravity in this case
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
so finally time required to reach maximum height t = U/g = 24/32 =3/4 second
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
so u can do this on remaining planets
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
So Mercury would be 2 seconds right
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
next maximum height reached by the ball
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
.85 seconds
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
for what?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
correct
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
do the remaining later lt's go for the next question
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
Wait how do you find the maximum height?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
for maximum height reached substitute t = U/g in h(t)
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
You said it was t = U/a?
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
So I substitute the time required to reach maximum height into the quadratic functions?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
u will get h(t) = U^2/2g + ho
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
So the answer to the first one would be 27 ft right
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
now on earth surface, h = (24)^2/2(32) +3 = 12
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
ya right 27.32
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
now u can do the remaining let's move to the next question
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
time required to return to planet's surface
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
if the ball is projected from the ground time taken to reach the maximum height = time taken to reach the ground from maximum height
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
So it's the same answer?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
wait this is some what lengthy
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
now, time taken by the ball to reach the hand of the person is, U/g + U/g = 2U/g
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
but the ball should reach the ground
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
so it reaches the hands of the person with a velocity of 24 feet per second
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
now it should reach the ground by traveling 3 feet
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
So 27 feet
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
no how do u say that?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
u der?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
now h(t) = (1/2)g(t^2) + Ut 3 = 16t^2 + 24t
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
t = 0.116 seconds
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
now add this with 2U/g
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
hope u understood
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
There's a negative in front of (1/2)g And how did you get 3?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
that is given because there the ball is projected vertically upward here the ball is coming down
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
it should travel 3 feet downward to reach the ground
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
\[h(t)=\frac{ 1 }{ 2 }g t ^{2}+v _{0}t +h _{0}\] So you got the 3 from Earth's quadratic function right?
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
So does that mean they all travel 3 feet downward?
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
but the ho for each of them is 3
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
the ball should travel 3 feet downward after it reaches the hands of the person ,to reach the ground
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
So how long do the other ones travel?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
same 3 feet
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
So the other ones travel 3 feet downward?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
dw:1355813142620:dw
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
i think now u can understand clearly
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
Ok thank you
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.2
welcome
 one year ago

Mathhelp346 Group TitleBest ResponseYou've already chosen the best response.0
Mercury's time would be 4 seconds?
 one year ago
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