projectile/assignment help (1 Viewer)

[twistedrebel]

So for my assignment i had to find initial velocity, final velocity, range, maximum height and time of flight. by throwing a projectile

I found the range, buy measuring how far the ball went, and the time by recording it. I know that initial velocity is 0. I could find the final velocity and maximum height using the formulas, but i ahve one problem.

I threw the ball via my hand, which created 43 degree angle (i worked it out by the photo taken), the height was 185c from the floor, but i measure the time of flight when it hit the floor. So i am wondering whats the best way to calculate all the things i need, taking into account that i threw it from 185cm height to the floor.

 

[annabackwards]

Make the origin the point where the ball was thrown and the ground -Xcm where X is whatever the height of the ball above the ground is.

Oh and the initial velocity could not be zero unless you dropped it. Seeing as you said you "threw" it, the intial velocity could not have possibly been zero. Unless you are starting the time when the ball reaches maximum height...

You can however use some of your measurements to find out what your initial velocity is though

EDIT: sikhman beat me XD

 

[annabackwards]

Easiest way is to use the 3U formulas (find them on hyperphysics if you can't be bothered to derive them)

But he should practise with his physics formulas. It's not that hard anyway with the physics equations

 

[annabackwards]

Yeah but I'm lazy

Fair enough

@OP - reply to this thread if you need more help

 

[twistedrebel]

how will i work out the initial velocity? i m failling at this calculation part :S

 

[annabackwards]

how will i work out the initial velocity? i m failling at this calculation part :S

Make sure you get the time it takes to get max range - not when it hits the floor but when it reaches the other side of the parabola in line with where the ball was initially launched.

Also make sure you know what the max height of the ball was and it'd make life easier for you if you could find the time it took to get to that height

 

[twistedrebel]

shady helped me with this

Distance= 4.36m (measured)
time=1.01sec

find U (x) and v(x)
using speed= distance /time
u(x) 4.36/1.01
u(x) = 4.32 m/s
v=u + at
v(x)= 4.32 + 0x1.01
v(x)= 4.32
Finding the initial Velocity of the Projectile
U(x) = Vcos(angle)
4.32= Vcos43
v=4.32/cocs43
v=5.9m/s at 43 degrees above the horizontal.
Thats all of the direction finished
a(y)=-9.8
t(y)=1.01
R(y)=2.42
Finding the Initial Velocity in the "y" direction
U(y) =V sin(angle)
u(y)=4.9 x sin (43)
u(y)=3.34
Finding range in "y" direction
choose point V(y)=0, maximum of the projectile
Now v(y)=U(y)+at
0=3.34-9.8xt
t=3.34/9.8 =0.34secs (time taken to get to them highest point)
Now
R(y)=ut+1/2xat^2
R(y) 3.34x.34-4.9x(.34)^2
R(y)= 0.57m from the point of fire to maximum height.
so height from groundis 1.85+0.57=2.42M
Finding the final velocity in 'y' direction
we split it into 2 projectiles, this part being from the maximum height to the ground
make the angle=0 degrees and u(y) =0
now to find how this is, (in secs) we simply subtract the time we found before (for first part of the projectile, which was the starting position to the maximum height) from the overall time.
so time for 'second projection' from maximum height to the ground is =1.01-.34=0.67secs
v(y)=u(y)+at
keep in mid u(y)=0
v(y) =0+9.8x0.67 =6.566m/s (we choose gravity to be positive)
Final velocity is using 2 vectors and pythatgroas theorem to work it out.
(4.32)^2+(3.43)^2=5.516m/s

If someone could check all calculations are correct, appreciated. (please if you can do calculations, to make sure i did not make a mistake anywhere)

 

[strasiotto]

I'm sure I've missed something, but where did you get the time from?

 

[strasiotto]

cheers lol. I thought there was some equation I'd missed

 

[zombie omie jay]

m8 all you needed was to time how long it took to reach its range
and you can calc the speed