got some velocity question with projectile (1 Viewer)

[wolf7]

a ball is kicked at a velocity of 40m/ss at 60 degrees above the horizontal

A) When is the ball 25m above the ground?
b) when and where will the ball strike the ground again
c)if the ball is to pass over a 25m high, between what two distnace from the launching point can the barrier be placed

 

[Spoz]

The HSC is over for me but not my friends, so solving physics problems on the internet will have to settle for amusement

a)
Vertically the velocity is:
40Sin60 = 34.64m/s

Using s = ut + 1/2at^2

25 = 0 + 1/2 * 9.8 * t^2 *** Edit: u = 34.64 not zero. I am an idiot, and the next post does things properly. There has to be a way to do it without the quadratic formula though.

Rearranging:
t^2 = 25/4.9

t = 2.2587s

But its 25m off the ground on the way down TOO, which I'll do in part B and C. In an exam you would need to do all those steps in part A of the question, but I am lazy.

B) Ball reaches maximum height when vertical velocity is zero.

v = u + at
0 = 34.64 + 9.8t
t = 34.64/9.8
t = 3.535

Ball lands after 2*3.545s so the time of flight is 7.069s

Horizontally the velocity is
40Cos60 = 20m/s

s = ut + 1/2at^2
s = 20 * 7.069+ 0
= 141.4m

thus the ball will strike the ground 141m from where it is kicked, after 7s.

C) You need to work out when the ball is higher than 25m.
You (I) did this in part A.
The ball is higher than 25m After 2.2587s and before (7.096-2.2587) s

Ie between 2.2587s and 4.8103s

Horizontally
s = ut + 1/2at^2
s = 20*2.2587 + 0
s = 45.17m

and

s = 20 * 4.8103
s = 96.2m

So the fence could be placed anywhere between 45m and 96m away.

There might be slight errors somewhere but you get the idea of how to work it out

 

[wolf7]

for question a, why is the initial velocity 0

 

[Riviet]

for question a, why is the initial velocity 0

It's not 0, otherwise the ball wouldn't be moving lol. It would still be on the ground xD.
a) Use u=34.64 instead of 0 lol, and solve for t.
Since the ball goes up, passing the point where it is 25m above ground, reaches it's max height, then comes back down passing the 25m point above ground again, there are two times. Hence the quadratic that you need to solve.
4.9t^2 - 34.64t +25=0
Using the quadratic formula, t=0.82s, 6.25s
b) First we find time of flight up to where ball reaches max height
v=u+at
t=(v-u)/a
=(0-34.64)/(-9.8)
=3.53s
.: total time of flight is 2x3.53= 7.06s
Now let's find the horizontal u:
u_horizontal=40cos60=20m/s
Now using total time, t=7.06
distance=speedxtime lol, simple equation
=20x7.06
=141.2m, which is how far the ball flies horizontally
c) From a), we found that the ball is at a height of 25m above the ground at t=0.82 and 6.25.
And in b) we found that the horizontal speed is 20m/s
.: Distance that barrier/fence (whatever u wanna call it lol) must be placed from starting position = 0.82x20 =16.4m for when it goes up
and 6.25x20=125m for when it comes back down.
That's it.

 

[Spoz]

Fuck a duck. All that time and I screwed up in the second line. Probably a good thing I dont do physics anymore

 

[wolf7]

can u show us the quadratic formula and how to work it out

because, i dont have a clue which formula ur using

 

[richz]

dont u do maths???

t=(-b+-sq root (b²-4ac))/2a

coeff of t² is a
coeff of t is b
constant is c

note: +- is plus or minus

 

[Riviet]

can u show us the quadratic formula and how to work it out

because, i dont have a clue which formula ur using

From the quadratic 4.9t² - 34.64t +25=0
a=4.9, b=-34.64, c=25.
Now the quadratic formula is:
t=-b+/-sqrt(b²-4ac) (sqrt=square root of)
2a
=34.64+/-sqrt[(-34.64)²-4(4.9)(25)
2x4.9
=34.64+/-sqrt(709.9)
9.8
.: t=(34.64+26.6)/9.8 or t=(34.64-26.6)/9.8
t=6.25s, 0.82s