Deriving the Equation of Motion for an Object Thrown Vertically Upwards

[Deimantas]

Homework Statement

From a height of 18 meters above ground, an object is thrown vertically upwards at a speed of 30m/s. Find the equation where height h depends on time t. Find the maximum height of the object.

Homework Equations

h=h₀+v₀t-(gt^2)/2

The Attempt at a Solution

The equation of motion of an object thrown vertically upwards is h=h₀+v₀t-(gt^2)/2
By inserting h₀=18 and v₀=30 we get
h=18+30t-(gt^2)/2, which is the correct answer. When t=3, h is max at 63.
Now my problem here is that instead of finding the aforementioned equation of motion in a book, i was supposed to create it myself, using differential equation methods. How?

 

[HallsofIvy]

Once the ball have been thrown up it has accelration -g, due to gravity. Acceleration is the derivative of velocity and velocity is the derivative of distance (height here).

So your differential equations problem is d^2h/dt^2= -g with initial values h(0)= 18, h'(0)= v(0)= 30.

 

[Dick]

Homework Statement

From a height of 18 meters above ground, an object is thrown vertically upwards at a speed of 30m/s. Find the equation where height h depends on time t. Find the maximum height of the object.

Homework Equations

h=h₀+v₀t-(gt^2)/2

The Attempt at a Solution

The equation of motion of an object thrown vertically upwards is h=h₀+v₀t-(gt^2)/2
By inserting h₀=18 and v₀=30 we get
h=18+30t-(gt^2)/2, which is the correct answer. When t=3, h is max at 63.
Now my problem here is that instead of finding the aforementioned equation of motion in a book, i was supposed to create it myself, using differential equation methods. How?

You want to start from using that the derivative of the velocity is equal to the acceleration. So v'(t)=(-g). Integrate both sides to get v(t). Determine the constant of integration. Then v(t)=h'(t), so integrate again to get h(t)

 

[Deimantas]

Thanks for helping me out :)