Question

a boy throws stone with a speed of 12m/s if he releases them at a height 1.25m the maximum distance from his feet where they land is?

Answer 1

i got the answer as 14.4m but the orginal answer is 15.6m...

Answer 2

i used Rmax=u^2/g

Answer 3

So he threw the stone vertically up?

Answer 4

no

Answer 5

If I use the equation \[d = \frac{v^2}{g}\sin(2\theta)\] and set theta to 45, then I get 14.4

Answer 6

i think we have to add 1.25m also

Answer 7

right, but that gives 15.65 instead of just 15.6

Answer 8

can we make a approximate answer

Answer 9

I don't think so. I think the 15.6 answer came from some working out :)

Answer 10

I used \[y = v_0\sin(\theta)t - \frac{1}{2}gt^2\]and solved for t with y = -1.25, v_0 = 12, g = 10, and theta = 45 degrees. Then I took the positive solution and plugged it into\[ x = v_0\cos(\theta)t\]to get something like 15.55 m

Answer 11

something like \[x = \frac{3(\sqrt{194} + 12)}{5}\text{m}\]

Answer 12

let me solve this and i will inform u.......

Answer 13

wat is the t?

Answer 14

After solving that first equation, I got a negative t, and\[t = \frac{\sqrt{97} + 6\sqrt{2}}{10}\]