- #1
vizakenjack
- 57
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An object fired at an angle of 30° above the horizontal takes 2.5 s to travel the last 12 m of its vertical distance and the last 10 m of its horizontal distance. With what speed was the object launched? (Note: This problem does not stipulate that the initial and final elevation of the object has to be the same!)
Solution:
First, an initial velocity in x direction (air friction is ignored, I guess, so velocity in x direction stays the same) is to be found.
The object traveled in x direction - 10 meters.
And the formula for horizontal distance is: d = vt
10 = v*2.5s
v = 4m/s
that's the x component of velocity (Vx)
Knowing initial Vx and an angle, we can find the y component of the velocity (Vy)
Tangent is opposite/adjacent. We need to find opposite, which corresponds to Vy.
so: tan(30°) * Vx (adjacent) = Vy
Vy = 2.309 m/s
Now the initial velocity needs to be found. Using Pythagoras theorem: sqrt(4^2 m/s + 2.309^2 m/s) =
4.618602 m/s
The answer is: 4.618602 m/s
But my question is, why doesn't the solution work if instead of finding initial horizontal velocity, initial vertical velocity is to be found first?
See for yourself:
So I see 12 meters is the max height reached by an object, right?
Formula for height in projectile motion is: h = V2y/2g -> 12=V2y/2g
Vy = 15.3362 m/s
It's already wrong!
Why??
I'm also kinda confused whether I should use this formula:
d = V*t + 1/2 * gt2
Anyway, using that kinematic equation yields: 17.05 m/s for Vy
Which is also wrong...
Solution:
First, an initial velocity in x direction (air friction is ignored, I guess, so velocity in x direction stays the same) is to be found.
The object traveled in x direction - 10 meters.
And the formula for horizontal distance is: d = vt
10 = v*2.5s
v = 4m/s
that's the x component of velocity (Vx)
Knowing initial Vx and an angle, we can find the y component of the velocity (Vy)
Tangent is opposite/adjacent. We need to find opposite, which corresponds to Vy.
so: tan(30°) * Vx (adjacent) = Vy
Vy = 2.309 m/s
Now the initial velocity needs to be found. Using Pythagoras theorem: sqrt(4^2 m/s + 2.309^2 m/s) =
4.618602 m/s
The answer is: 4.618602 m/s
But my question is, why doesn't the solution work if instead of finding initial horizontal velocity, initial vertical velocity is to be found first?
See for yourself:
So I see 12 meters is the max height reached by an object, right?
Formula for height in projectile motion is: h = V2y/2g -> 12=V2y/2g
Vy = 15.3362 m/s
It's already wrong!
Why??
I'm also kinda confused whether I should use this formula:
d = V*t + 1/2 * gt2
Anyway, using that kinematic equation yields: 17.05 m/s for Vy
Which is also wrong...
Last edited: