Physics help?
I can't seem to remember how i did this problem back when i did the homework. Could someone remind me how to find the components without using the angle that it was projected from? Any help is appreciated. Thanks~
A projectile launched from the ground lands 120 m away at t = 3 s later. Use g = 10 m/s2.
What is the x-component of the initial velocity vx0?
What is the y-component of the initial velocity vy0?
For the x-component, i did 120/3 = 40 m/s to find the X. But i'm not sure if this was the correct method. For the y, i can't seem to figure out how to find the component...
February 23, 2014
6 Comments • Newest first
Because this is projectile motion, your horizontal component of initial velocity remains constant.
So x = 120m, t, = 3s, vx0 = 120m/3s = 40 m/s, so you're right on that.
Because the projectile started from the ground then went to a maximum height then went down, the time it took to reach its maximum height from the ground is the same amount as it takes to get from the max height to the ground. In other words, time up = time down (provided air resistance is negligible, which I'm sure it is in this context)
Total time = 3s = time up + time down, so time up = 1.5s
vf = vi + at
at the maximum height in a projectile motion, velocity is 0 m/s, in this case vf is the velocity at the maximum height, and we are solving for vi (initial vertical velocity)
vi = vf - at
vi = 0 m/s - (-10 ms^2)(1.5s) [gravity is a downward acceleration so it is negative]
vi = 15 m/s
[quote=Phong128992]Some teachers don't really care if it's that or 10. Physics is more about getting the concepts not using the numbers. Sorry that you had a teacher that was strict on that.[/quote]
Idk.. every teacher i know who knows anything about physics always remembered that it was supposed to be kept at 9.8 and not 10
[quote=Gmayn]Yes, 40 m/s is correct for horizontal velocity -- I believe. Vertical component would be v = v0 + at.
0 = v0 + (-10)(1.5) <-- time is half of total time when the vertical velocity is 0.
v0 = 15 m/s
Correct me if I'm wrong, anyone.[/quote]
You can also use 0 = 3vsin(theta) + (.5)(-10)(3)^2, but yeah, that works too.
[quote=Thiefy996]I know it would give a messier result, but I cringed when I saw g=10m/s^2
My teachers were always really stern on teaching that [b]IT'S 9.81 m/s^2 AND DON'T YA FORGET IT[/b].If there's one thing I learned from physics class...[/quote]
Some teachers don't really care if it's that or 10. Physics is more about getting the concepts not using the numbers. Sorry that you had a teacher that was strict on that.
Just take the derivative or something.
Yes, 40 m/s is correct for horizontal velocity -- I believe. Vertical component would be v = v0 + at.
0 = v0 + (-10)(1.5) <-- time is half of total time when the vertical velocity is 0.
v0 = 15 m/s
Correct me if I'm wrong, anyone.