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

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Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0Activity 1 is the first picture

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2what is the velocity of projection???

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2do u know the equations of motion??

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2time taken to reach the maximum height t = u/g

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

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

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2no do u know the formula V = U + gt

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2u r of which standard??

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2do u know about displacement,velocity,acceleration etc

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2okay 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

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2now ur first question maximum height of the ball

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2at maximum heicht, final velocity is zero

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2so, 0=U+at t = U/a

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2here acceleration is nothing but acceleration due to gravity

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

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0What do you mean acceleration is nothing?

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2acceleration is equal to acceleration due to gravity in this case

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

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2so u can do this on remaining planets

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0So Mercury would be 2 seconds right

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2next maximum height reached by the ball

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2do the remaining later lt's go for the next question

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0Wait how do you find the maximum height?

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

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0You said it was t = U/a?

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0So I substitute the time required to reach maximum height into the quadratic functions?

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

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0So the answer to the first one would be 27 ft right

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

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2now u can do the remaining let's move to the next question

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2time required to return to planet's surface

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

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0So it's the same answer?

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2wait this is some what lengthy

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

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2but the ball should reach the ground

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2there 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

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

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2now it should reach the ground by traveling 3 feet

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2no how do u say that?

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

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2now add this with 2U/g

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

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

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2it should travel 3 feet downward to reach the ground

Mathhelp346
 2 years ago
Best 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?

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0So does that mean they all travel 3 feet downward?

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0but the ho for each of them is 3

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

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0So how long do the other ones travel?

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0So the other ones travel 3 feet downward?

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1355813142620:dw

bhaskarbabu
 2 years ago
Best ResponseYou've already chosen the best response.2i think now u can understand clearly

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.0Mercury's time would be 4 seconds?
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