Math-Based-Physics question.

Question

Math-Based-Physics  question.

yolomcswagginsggg · Answer

9 + 10?

anonymous · Answer

19

yolomcswagginsggg · Answer

21.

idku · Answer

A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s.
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idku · Answer

STOP IT, GOD DAMN IT !

yolomcswagginsggg · Answer

xD

idku · Answer

`PART 1:`
1)  What is the initial speed of the cannonball?
$\sqrt{33^2(m/s)+25^2(m/s)~}=41(m/s)$

idku · Answer

`PART 2:`
2)  What is the initial angle θ of the cannonball with respect to the ground?`PART 1:`
$\theta=\tan^{-1}\left(25/33\right)\approx37.15$

idku · Answer

`PART 3`
3)  What is the maximum height the cannonball goes above the ground?

idku · Answer

$\Large v^2=u^2+2ax$
(NOTE: x is vertical displacement)

Ok, so I know that v=0 when the ball reaches maximum height.
My initial vertical velocity is 33m/s.
Acceleration is -9.81a

$\Large 0^2=33^2+2(-9.81)x$
$\Large 33^2=2(9.81)x$
$\Large x=55.5$

and that is wrong...

idku · Answer

I just want to do it a plain way using some familiar formula

idku · Answer

oh, it should be 25 not 33

idku · Answer

because we are computing vertical displacement we must use vertical velocity

idku · Answer

$\Large v^2=u^2+2ax$
(NOTE: x is vertical displacement)

Ok, so I know that v=0 when the ball reaches maximum height.
My initial vertical velocity is 25m/s.
Acceleration is -9.81a

$\Large 0^2=25^2+2(-9.81)x$
$\Large 25^2=2(9.81)x$
$\Large x=31.8$

and that is CORRECT!

idku · Answer

Okay, I am getting through it with a painful struggle-:(:)

ganeshie8 · Answer

good, nothing to feel bad about
these things take time

idku · Answer

Like nothing else in this world.... :)
But, true that

idku · Answer

Well, the fact that my teacher is awful gives me more credit

idku · Answer

4)  How far from where it was shot will the cannonball land?

idku · Answer

$\Large x = ut + \dfrac{1}{2}at^2$

$\Large 31.8 = 25t - \dfrac{9.81}{2}t^2$

t = 2.44229
t = 2.65455

then I can use:
Horizontal displacement = time × initial horizontal velocity

my problem now is which time to take

ganeshie8 · Answer

you should not get multiple "t" values
the vertex is a "single" point 
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