Ok: A hammer free falls 80 m, disregarding air resistance. What’s the velocity when it hits the ground.
Initial Velocity (Vi) = 0 m/s
Final Velocity (Vf) = Unknown
Acceleration due to gravity (g) = -9.8 m/s[sup]2[/sup]
Displacement (delta x) = 80 m
Vf [sup]2[/sup] = Vi[sup]2[/sup] + (2)(g)(delta x)
Vf [sup]2[/sup] = 0[sup]2[/sup] + (2)(-9.8)(80)
Vf [sup]2[/sup] = 0 + -1568
I’m not sure what the square root of -1568 is, but how can a velocity be imaginary? Where did I mess up?
You’re using opposite axes for acceleration and gravity. In other words, if acceleration is -9.8m/s[sup]2[/sup] (’-’ means down), then displacement should be -80 m (also down).
I didn’t just do your homework for you, did I?
I’ll admit that was a homework question. However, it was the concept had been holding me up on several questions and with that knowledge I wont make that mistake again. Thanks robot!
I always was taught that D= 1/2AT[sup]2[/sup]. This would be 80m= 4.85T[sup]2[/sup], making T approximately 4.06 seconds, multiplied by 9.7 M per sec[sup]2[/sup], equals 39.39 mps.
Qwerty:
First of all, we generally use the y-axis for free fall problems. Secondly, note that displacement is the final position minus the initial position. As Robot stated, you want to be consistent in your choice of axis directions. Pick “upward” to be the direction of the +y-axis. Since the acceleration due to gravity is downward, a[sub]y[/sub] = -9.8m/s[sup]2[/sup]. Similarly, your initial position is y[sub]0[/sub] = 80 m, and your final position is y = 0. Taking delta(y) = y - y[sub]0[/sub] = 0 - 80 m, we get that the displacement is -80 m.
Lumpy, your equation is only valid if the initial vertical velocity (v[sub]y0[/sub]) is zero.
>> You’re using opposite axes
Wasn’t it hammers we was talking about? 