# Ball thrown upwards

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• Sep 20th 2010, 03:20 PM
HelloWorld2
Ball thrown upwards
A ball weighing .15kg is thrown upwards at a rate of 20m/s from a building 30m high. Neglecting air resistance, formulate a differential equation govering the ball's motion, solve it, and use the solution to determine the maximum height of the ball.
• Sep 20th 2010, 03:24 PM
mr fantastic
Quote:

Originally Posted by HelloWorld2
A ball weighing .15kg is thrown upwards at a rate of 20m/s from a building 30m high. Neglecting air resistance, formulate a differential equation govering the ball's motion, solve it, and use the solution to determine the maximum height of the ball.

Motion up: $\displaystyle \frac{d^2 x}{dt^2} = -g$ subject to the boundary conditions that at t = 0, $\displaystyle \frac{dx}{dt} = 20$ and $x = 0$. (SI units used).
• Sep 20th 2010, 03:36 PM
undefined
Quote:

Originally Posted by mr fantastic
Motion up: $\displaystyle \frac{d^2 x}{dt^2} = -g$ subject to the boundary conditions that at t = 0, $\displaystyle \frac{dx}{dt} = 20$ and $x = 0$. (SI units used).

This treats the top of the building as x = 0; in reference to the ground we can put x = 30.
• Sep 20th 2010, 03:43 PM
HelloWorld2
So we just find an explicit expression for velocity, then integrate it, and apply the fact that s(0)=30 for the contstant. Then plug in the t that makes v=0?
• Sep 20th 2010, 06:17 PM
mr fantastic
Quote:

Originally Posted by HelloWorld2
So we just find an explicit expression for velocity, then integrate it, and apply the fact that s(0)=30 for the contstant. Then plug in the t that makes v=0?

Yes.