Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

Activity 1 is the first picture

what is the velocity of projection???

Does that help?

yep

do u know the equations of motion??

What do you mean?

okay leave it

Ok

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

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

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

no do u know the formula
V = U + gt

No

u r of which standard??

class of study?

8th grade?

do u know about displacement,velocity,acceleration etc

Yes

Ok

now ur first question
maximum height of the ball

at maximum heicht,
final velocity is zero

Ok

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

here acceleration is nothing but acceleration due to gravity

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

so finally t =U/g

What do you mean acceleration is nothing?

acceleration is equal to acceleration due to gravity in this case

Ok

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

Ok

so u can do this on remaining planets

So Mercury would be 2 seconds right

2.02

next maximum height reached by the ball

.85 seconds

for what?

Venus

correct

Ok

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

Ok

Wait how do you find the maximum height?

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

You said it was t = -U/a?

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

ya

Ok

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

Ok

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

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

ya right 27.32

Ok

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

Ok

time required to return to planet's surface

Ok

So it's the same answer?

no

wait this is some what lengthy

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

Ok

but the ball should reach the ground

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

now it should reach the ground by traveling 3 feet

So 27 feet

no how do u say that?

u der?

Yes

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

t = 0.116 seconds

now add this with 2U/g

hope u understood

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

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

it should travel 3 feet downward to reach the ground

ya

So does that mean they all travel 3 feet downward?

no

but the ho for each of them is 3

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

So how long do the other ones travel?

same 3 feet

So the other ones travel 3 feet downward?

|dw:1355813142620:dw|

Ok

i think now u can understand clearly

okay?

Ok thank you

welcome

:)

Mercury's time would be 4 seconds?