Launching from a height.

Step 1 of 2

How long in the air

The cannon now stands on a tower 20 m high. The ball still goes up and comes down as before, but at the end it keeps falling all the way to the field, another 20 m below.

The formula $t = \dfrac{2 v\sin\theta}{g}$$t =$$\dfrac{2 v\sin\theta}{g}$⁠ no longer fits: it only holds when the ball lands at the height it was launched from. Write the vertical position as a function of time and find when it reaches the field:

$y(t) = v\sin\theta\, t - \tfrac12 g t^2$$y(t) =$$v\sin\theta\, t - \tfrac12 g t^2$⁠

Step 2 of 2

Is 45 degrees still the best

The speed is now 15 m/s and the cannon is still on the tower. Launch angle and Launch speed are free to change: try a few angles and compare the distances the Scene marks.