Object Dropped From 500m: Solved with Differential Equations

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An object dropped from a height of 500 meters can be analyzed using differential equations, specifically applying the law of free fall with an acceleration of 9.81 m/s². By integrating the acceleration equation, the velocity can be expressed as V(t) = 9.81t, with the initial velocity set to zero. The displacement equation is derived as D = 4.9t², allowing for the calculation of time when D equals 500 meters. Substituting this time back into the velocity equation provides the final speed upon impact. This method effectively combines differential equations with basic physics principles to solve the problem.
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an object is dropped from a height of 500m. when will object reach the ground level and with what speed?

important: the solution must be by: Differential Equations.
 
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Okay, so what equations has your lecturer given you that might relate to this problem?
 
the Lecturer gave us a question without any equations or laws - just said to me: do it by using law of free fall !

by the way- our subject is MATH
 
Your best approach is to use acceleration equals 9.81 meters per second per second. From there, you can integrate your equation until you get a velocity equation and finally the displacement equation. All you need to know is acceleration and once you integrate using initial velocity as zero (or v(0)=0), you can set the displacement equation equal to 500. You can also use basic physics kinematic equations to find the time, and then plug in the time to your velocity equation to find the final velocity.

Note: t=time

A(t)=9.81

V(t)=9.81t+C, but we know C=0 because V(0)=0

D=4.9t2
 
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