# Baffling simple physics question **URL:** \ **Category:** Factual Questions\ **Created:** [September 18, 2001, 1:07am UTC](https://boards.straightdope.com/t/baffling-simple-physics-question/82298 "2001-09-18T01:07:00Z")\ **Posts on this page:** 6\ **Page:** 1
**Author:** ![Qwertyasdfg](https://avatars.discourse-cdn.com/v4/letter/q/ecc23a/32.png) [@Qwertyasdfg](https://boards.straightdope.com/u/Qwertyasdfg)\ **Post date:** [September 18, 2001, 1:07am UTC](https://boards.straightdope.com/t/baffling-simple-physics-question/82298/1 "2001-09-18T01:07:00Z")
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? ---
**Author:** ![Robot\_Arm](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robot_arm/32/18280_2.png) [@Robot\_Arm](https://boards.straightdope.com/u/Robot_Arm)\ **Post date:** [September 18, 2001, 1:15am UTC](https://boards.straightdope.com/t/baffling-simple-physics-question/82298/2 "2001-09-18T01:15:02Z")
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? ---
**Author:** ![Qwertyasdfg](https://avatars.discourse-cdn.com/v4/letter/q/ecc23a/32.png) [@Qwertyasdfg](https://boards.straightdope.com/u/Qwertyasdfg)\ **Post date:** [September 18, 2001, 1:21am UTC](https://boards.straightdope.com/t/baffling-simple-physics-question/82298/3 "2001-09-18T01:21:02Z")
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! ---
**Author:** ![Lumpy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lumpy/32/446_2.png) [@Lumpy](https://boards.straightdope.com/u/Lumpy)\ **Post date:** [September 18, 2001, 2:59am UTC](https://boards.straightdope.com/t/baffling-simple-physics-question/82298/4 "2001-09-18T02:59:54Z")
I always was taught that D= 1/2A_T[sup]2[/sup]. This would be 80m= 4.85_T[sup]2[/sup], making T approximately 4.06 seconds, multiplied by 9.7 M per sec[sup]2[/sup], equals 39.39 mps. ---
**Author:** ![robby](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robby/32/11048_2.png) [@robby](https://boards.straightdope.com/u/robby)\ **Post date:** [September 18, 2001, 3:58am UTC](https://boards.straightdope.com/t/baffling-simple-physics-question/82298/5 "2001-09-18T03:58:12Z")
**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. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [September 18, 2001, 4:23am UTC](https://boards.straightdope.com/t/baffling-simple-physics-question/82298/6 "2001-09-18T04:23:52Z")
\>\> You’re using opposite axes Wasn’t it hammers we was talking about? 😉