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 May 13th, 2015, 10:11 AM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Free Fall Motion Under Gravity. A particle is projected vertically upward from ground level with a speed of 24.5m/s. Find a)The greatest height reached b) the time that elapses before the particle returns to the ground. My attempt for question a. $\displaystyle a=-9.8$ and $\displaystyle u=24.5$ $\displaystyle v^2-u^2=2as$ $\displaystyle 0^2-24.5^2=2(-9.8)s$ $\displaystyle s=30.6m$ For question b. $\displaystyle v=u+at$ $\displaystyle 0=24.5+(-9.8)t$ $\displaystyle t=2.5s$ But the answer for question b is 5s. Where's my mistake?
May 13th, 2015, 11:38 AM   #2
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 Originally Posted by jiasyuen A particle is projected vertically upward from ground level with a speed of 24.5m/s. Find a)The greatest height reached b) the time that elapses before the particle returns to the ground. My attempt for question a. $\displaystyle a=-9.8$ and $\displaystyle u=24.5$ $\displaystyle v^2-u^2=2as$ $\displaystyle 0^2-24.5^2=2(-9.8)s$ $\displaystyle s=30.6m$ For question b. $\displaystyle v=u+at$ $\displaystyle 0=24.5+(-9.8)t$ $\displaystyle t=2.5s$ But the answer for question b is 5s. Where's my mistake?
A common misconception. When we talk about the speed of the object when it hits the ground we are actually talking about the speed it has just before it hits the ground. So v is not 0.

Try $\displaystyle y = y_0 + v_0 t - (1/2)gt^2$. It is launched from the ground so $\displaystyle y_0 = 0$, and it hits the ground at the end, with y = 0. Now solve for t.

-Dan

Last edited by skipjack; May 14th, 2015 at 02:51 AM.

 May 13th, 2015, 06:08 PM #3 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Is my question a correct ?
 May 13th, 2015, 07:38 PM #4 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timey-wimey stuff. Your answer for part a is correct. For part b you tell me. Using my suggestion you need to solve $\displaystyle 0 = 24.5t - 4.9t^2$. What's t? -Dan
 May 14th, 2015, 02:00 AM #5 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 t is 5 second.
 May 14th, 2015, 02:56 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,104 Thanks: 1907 For question b, $\displaystyle v=u+at$ $\displaystyle -24.5=24.5+(-9.8/\text{s})t$ $\displaystyle t=5\text{s}$.
 May 14th, 2015, 03:16 AM #7 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,133 Thanks: 719 Math Focus: Physics, mathematical modelling, numerical and computational solutions Consider the following situations, all of which are relevant for a ball colliding with the floor: Situation 1: Free-fall. A ball is flying through the air and is subject to the force of gravity (and drag if you want to include it). Situation 2: Collision. A ball hits the ground. The exact moment it hits the ground, a force is exerted on the ball by the floor, in addition to gravity. The ball will decelerate and its motion will be affected by the new force. Typically the ball will deform and some energy is stored as elastic potential energy, depending on the substance the ball is made from. Situation 3: Equilibrium. If the ball doesn't bounce, it will eventually settle into an equilibrium state where the contact force from the floor balances the weight of the ball. Now, let's look at some of the parameters for each situation: Situation 1 (ignoring drag): Forces: weight only acceleration = constant motion: SUVAT equations can be applied Situation 2: Forces: weight $\displaystyle \neq$ contact force from floor acceleration = not necessarily constant (often approximated to be constant) as ball deforms in shape. Usually large. motion: more complex... cannot use SUVAT. Usually requires solving an ODE or using approximations. Situation 3: Forces: weight = contact force from floor acceleration = 0 motion: none (v=0) So.... whenever you solve any of these problems using SUVAT equations, your situation is entirely associated with situation 1. Therefore, you cannot set v = 0 because that is only relevant once the ball/particle/whatever has achieved equilibrium, which is in situation 3, which only occurs after a (generally complicated) collision.

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