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maximum height above the ground that the ball reaches

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First order differential equation model: A ball with mass m kg is thrown upward with initial velocity 19 m/sec from the roof of a building 22 m high. Neglect air resistance.

(a) Find the maximum height above the ground that the ball reaches.

(b) Assuming that the ball misses the building on the way down find the time that it hits the ground.

Use g = 9.8 m/sec2. Round the values to one decimal place.

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Solution Preview

We'll need two first-order differential equations here (or one second-order equation).

Let h(t) be the height of the ball at time t, and
v(t) be the velocity of the ball at time t.

We have: h'(t) = v(t), velocity is the derivative of the position function,
v'(t) = -g = -9.8, acceleration is the first derivative of the velocity function. We have to put a minus at g, because the velocity vector is directed upward, and the acceleration acts in the opposite direction, ...

Solution Summary

The maximum height above the ground that the ball reaches is calculated.