Question

a lovesick lad wants to throw a bag of candy and love notes into the open window of his friends bedroom 10m above he throws the gift at 60 degrees to the ground:

at wht velocity should he throw the bag?

Answer 1

@ParthKohli

Answer 2

Let's assume that the velocity is \(x\). What is the vertical velocity of the bag of candy?

Answer 3

You know what "vertical velocity" means, yes?

Answer 4

yes i do know wht it means :P
we r not given any velocity

Answer 5

Yes, assume the velocity as a variable.

Answer 6

If the velocity is \(x\), then the vertical velocity is \(v\sin(60^{\circ})\)

Answer 7

Now, do you know what the max height is, in a projectile?

Answer 8

in general ?\or about that q;?

Answer 9

In general... we'll apply that to the question.

Answer 10

no

Answer 11

Let's derive that for a vertical motion.

Answer 12

\[v^2 = u^2 - 2gH\]Now, in vertical motion, the max height is when \(v = 0\).\[0^2 = u^2 - 2gH \Rightarrow H_{max} = \dfrac{u^2}{2g}\]

Answer 13

So, the max height is\[\dfrac{u^2}{2g}\]Now, what is the initial vertical velocity?

Answer 14

And what is the max height this projectile covers?

Answer 15

2*9.8

Answer 16

?

Answer 17

\[H_{\large max}=\dfrac{u^2}{2g}\]

Answer 18

\[H_{\rm max}= 10~ m\]\[u= x\sin(60^{\circ})\]Plug that into the formula.

Answer 19

i got 12

Answer 20

\[200 = \dfrac{3}{4}x^2\]

Answer 21

wht??

Answer 22

Hmm, let's solve it.\[10 = \dfrac{\left(x\dfrac{\sqrt{3}}{2}\right)^2}{2\times 9.8}\]

Answer 23

Get it?

Answer 24

nop, u just lost me with the above fraction

Answer 25

3/2
???????
where is that from?

Answer 26

\[u = x\sin(60) = x \dfrac{\sqrt3}{2}\]

Answer 27

how did u get 3/2?

Answer 28

\[\sin(60 ) = \frac{\sqrt{3}}{2}\]

Answer 29

oh ook....

Answer 30

i got 258?

Answer 31

Find the root of that.