- #1
Thunderer
- 10
- 0
I am not too sure whether I am doing this correctly.
The problem is basically a mini-cannon aimed at a 60 degree angle, hits a target on the ground 0.8m away. I am suppose to find the initial velocity.
This is what I did:
Cut the distance in half, assuming the mini cannon-ball will reach its maximum height at that point (and v = 0). Which is 0.4 m. I then use trig, and do the following: tan60 = y/0.4; y = 0.693m
vf^2 - vi^2 = 2a(xf-xi)
vf^2 - 0 = 2(-9.8m/sT=^2)(0.693m)
vf = 3.68 m/s from top of height to ground
so vi = 3.68 m/s
So this is the velocity of the y component?
Then:
tan 60 = (3.68 m/s) / x
xtan(60) = 3.68 m/s
x = 3.68/tan60 = 2.12 m/s
v(total)^2 = vy^2 + vx^2 = (3.68 m/s)^2 + (2.12 m/s)^2
v(total) = 4.25 m/s?
Is that correct? Did I do anything wrong?
[I also did this, but I don't think I need it.
vf = vi + at
0 = 3.68 m/s + (-9.8 m/s^2)(t)
t = 0.376 s ]
------
The problem is basically a mini-cannon aimed at a 60 degree angle, hits a target on the ground 0.8m away. I am suppose to find the initial velocity.
This is what I did:
Cut the distance in half, assuming the mini cannon-ball will reach its maximum height at that point (and v = 0). Which is 0.4 m. I then use trig, and do the following: tan60 = y/0.4; y = 0.693m
vf^2 - vi^2 = 2a(xf-xi)
vf^2 - 0 = 2(-9.8m/sT=^2)(0.693m)
vf = 3.68 m/s from top of height to ground
so vi = 3.68 m/s
So this is the velocity of the y component?
Then:
tan 60 = (3.68 m/s) / x
xtan(60) = 3.68 m/s
x = 3.68/tan60 = 2.12 m/s
v(total)^2 = vy^2 + vx^2 = (3.68 m/s)^2 + (2.12 m/s)^2
v(total) = 4.25 m/s?
Is that correct? Did I do anything wrong?
[I also did this, but I don't think I need it.
vf = vi + at
0 = 3.68 m/s + (-9.8 m/s^2)(t)
t = 0.376 s ]
------