Question

physics, please help.

can you tell me what you find from the square root solutions?
https://www.dropbox.com/s/5oc58leuwin8ej1/Screenshot%202015-12-19%2015.20.51.png?dl=0

Answer 1

first you need to work out the VERTICAL component of the velocity.

then you need the equation for height of the rock at any time t (this is a quadratic equation)

the height you are interested in is 20m (i.e. when it hits the plateau its height is 20m)

However - you can see from the diagram that is ALSO 20m high on th e way UP when it passes the height of the cliff.

So if you solve the height equation for h=20 you will find the 2 TIMES at which the rock is at 20m

the larger of those times is when it lands on the plateau

Answer 2

why not post your actual quadratic?

Answer 3

If you want to compare your result, I get a time of
\[ t= \frac{3\sqrt{3}+ \sqrt{11}}{2} \]

Answer 4

h = ut +at^2/2

20 = 30 cos(30) t -5t^2

-5t^2 +30 cos (30)t -20 =0

-t^2 +6cos(30)t -4 =0

-t^2 + 3sqrt(3) t -4 =0

a=-1
b=3sqrt(3)
c=-4

\[t = \frac{ -3\sqrt{3}\pm \sqrt{27-16}}{ -2 }\]

So I agree with your derivation except you have not allowed for the +/- for the complete solution

The first solution (when you use the - option ) is when the rock passes 20m height on the way UP.

Answer 5

I agree with Mr. Nood that separating this problem into two parts: one for vertical motion and one for horizontal motion, would be an appropriate approach. You'll need to develop an equation of motion for each part. Note that there is zero horizontal acceleration, but vertical acceleration stems from the pull of gravity.

focusing on vertical motion only,

1) what is the acc;n?
2) write the basic equation for the velocity as a function of time.
3) write an equation for vertical position as a function of time.

Answer 6

Christos is offline at the moment.

Answer 7

isn't the quadratic -4,9t^2 + 25,8t - 20 = 0 ?

Answer 8

@mathmale @MrNood @phi @IrishBoy123

Answer 9

sounds right

this is the one @phi also solved above, using g = -10
http://www.wolframalpha.com/input/?i=solve%3A+20+%3D+30*sin%2860%29t+%2B+%281%2F2%29%28-10%29t%5E2

Answer 10

Well yes - I have taken g = 10m/s^2 - but your figure is more accurate (g= 9.81m/s^2)
for the vertical velocity we have kept cos 30 = 2/sqrt3)

Answer 11

this here (-25,8-sqrt((-25,8)^2+4*(-4,9)*(-20)))/-9,8

evaluates to

5,9511609776

and with '+' it evaluates to

-0,6858548551

Answer 12

where am i wrong

Answer 13

Your equation and mine are very similar

using either set of figures the answer comes out at .95s and 4.25 s

check your solution of the quadratic

Answer 14

I justed put it on wolfram . doesnt give the same

Answer 15

or nearly the same

Answer 16

anyways

Answer 17

look inside the radical, its -4ac and a and c are negative

Answer 18

this is because my quadratic expression is -4,9t^2 + 25,8t - 20 = 0

Answer 19

30 sin 60 is 15 sqrt(3) or about 25.981

Answer 20

***
this here (-25,8-sqrt((-25,8)^2+4*(-4,9)*(-20)))/-9,8
***
that should have a minus sign ^
(-25,8-sqrt((-25,8)^2 - 4*(-4,9)*(-20)))/-9,8

Answer 21

yes

i tried saying that earlier🤔

Answer 22

thanks a lot