Question

plz view this ?

Answer 1

plz answer with steps questions 147 and 148

Answer 2

for Q (1470 ans is 0.66H and for (Q)148 the answer is c in the option

Answer 3

Just equate the time taken in the two cases.hope you can do it:)

Answer 4

i don't even understand the question
what topic is this?

Answer 5

kinematics

Answer 6

i have never heard of it

Answer 7

You need to ask these physics questions to the physics people.

Answer 8

cansomeone ans it plzz

Answer 9

@stormfire1 plzz help

Answer 10

sorry i have no idea what to do sorry = /???

Answer 11

I'm not sure I can help with these either...still looking though

Answer 12

ok try ur best

Answer 13

I do not understand this either = / it seem like there more to this assignment it look cut off in the bottom part when it say by X, = 3t and... what??

Answer 14

i have a guess, I'm not sure about this though

First situation - the acceleration is not yet reduced:
Time to go up
\[y= \frac{1}{2} g t^2\]\[H= \frac{1}{2} g t^2\]\[t=\sqrt(\frac{2H}{g})\]

Total time to go up and down :) = \[t=2\sqrt(\frac{2H}{g})\] (eq1)

Second situation - acceleration is reduced to g/2
Time to go up (take note that here acceleration is still g because g/2 is only the acceleration when the body goes down)
In this case, our y will be H-h
\[H-h=\frac{1}{2} g t^2\]\[t=\sqrt(\frac{2(H-h)}{g}\]

Time to go down
\[H=\frac{1}{2} (g/2) t^2\]\[t=2\sqrt(\frac{H}{g})\]

Total time up and down:
\[t=2\sqrt(\frac{H}{g})+\sqrt(\frac{2(H-h)}{g}\] (eq2)

Equate eq1 to eq2 and solve for h, you will get 0.6568H or 0.66H

Answer 15

thazzz

Answer 16

@marco: nice job...I don't see any problems with what you posted.

Answer 17

For qsn 148), i get the answer to be u2=(u1+u3)/2 ... But you mentioned earlier that the ans is c .. are you sure about it?

Answer 18

By the shortest time, are they referring to a trajectory that peaks close enough to assume y_max~H? If so, @marco26: agreed.