Firing a projectile, max height + flight time

future_vet

1. The problem statement, all variables and given/known data
A projectile is fired with an initial speed of 65.2 m/s at an anmgle of 34.4 degrees above the horizontal on a long flat firing range.
Determine the max heigh reached by the prkectile, the total time in the air and the total horizontal distance covered and the velocity of the projectile 1.50 seconds after firing.

2. Relevant equations and 3. The attempt at a solution
I am going to check the first parts now, and then continue once I am sure I am going in the right direction.

Vxo = vocos 34.5 = 65.2/0.824 = 79.1 m/s
Vyo = vosin34.5 = 65.2/0.566=115 m/s

t=vyo/g=11.7 seconds.
y=Vyot - 1/2gt^2= 675 meters maximum height.

And

y=yo + vyot - 1/2 gt^2
0 = 0 + 115t - 1/2 x 9.80 x t^2
t = 0 at initial point
t= 2 (115)/9.80 = 23.5 seonds.

Thanks!

future_vet

Actually here's the third part
Range:
x = 0. ax=0, vxo= 79.1 m/s and 11.7x2 = 23.4 which is the total time for the whole "trip".
x= 79.1 x 23.4 = 1850.94 meters

For the velocity after 1.50 seconds, I have no idea how to proceed... Some help would be greatly appreciated. Thanks!

future_vet

Do I use the same equation as in the previous part:
x= 79.1 x 23.4 = 1850.94 meters
but replace 23.4 by 1.50 seconds
=> 79.1 x 1.5 = 118.7 meters ?
How about the significant figures throughout the exercise?

Thanks!

HallsofIvy

?? You say Vxo= vocos 34.5 but then you DIVIDE by cos 34.5? Same for Vyo. Surely the x and y components of speed can't be greater than the speed itself!


Your values for vxo and vyo are wrong. Recalculate them- multiply by the trig functions- and recalculate the rest.

Again, your value for vxo is wrong. Recalculate.

"x" is not the velocity, it's horizomtal distance. The velocity is the <vx, vy> vector. Since, as you said before, there is no horizontal acceleration, vx is just vxo. vy is, of course, vyo- gt.

cristo

Check these components: V_x0=V₀*cos34.5; here you've divided!

(plus you've put 34.4 in the first line of the question!)

future_vet

!!! How brainless.

Well, here is the correction:
Same but multiplying instead gives us 5.15 m/s and 36.9 m/s.
t= 3.77 seconds.
Then for the distance following y=Vyot - 1/2gt^2= 69.5 meters maximum height.

future_vet

Then we have
y=yo + vyot - 1/2 gt^2
0 = 0 + 36.9t - 1/2 x 9.80 x t^2
t = 0 at initial point
t= 2 (36.9)/9.80 = 7.5 seconds.

future_vet

And finally for the range:
The time was 3.77 => 3.77 x 2 = 7.54 seconds for the whole trip
x= 5.15 x 7.54 = 38.8 meters

cristo

Check the x component of the initial velocity again. Other than that, I've not checked your arithmetic, but your method is correct.

Correct

x component of initial velocity is incorrect (as above)

Yes, but don't you want the velocity after 1.5 seconds?

future_vet

Oops... It is 34.5 degrees... =/

cristo

OK, I thought as much, and that that was a typo. Did you see my above post re a mistake in the x component of the initial velocity?

future_vet

Yes, thank you! It did look small. I can't believe I am making so many mistakes...

future_vet

Actually, yes I need the velocity not the distance.
Do I use:
vyo- gt = the velocity
=> = 36.9m/s - 9.80 m/s^2 x 1.50 seconds = 14.85 m/s ?

I think I spent too much time on this exercise, nothing is making sense anymore and I am mixing up my numbers...

cristo

That gives you the y component of the velocity at the time t=1.5. Then calculate the x component of the velocity at this point.

To calculate the velocity at this point you need to find (a) the speed (i.e. magnitude of the velocity) at this point, and (b) the angle the velocity makes to the horizontal.

So, you can write the velocity v at this point to be v=(vcosθ, vsinθ), and you need to solve equations for v and θ