Answer to Question #180471 in Classical Mechanics for Peter

Question

Hi, can anyone work this out for me? I have a compressed air cannon that fires a standard golf ball.

Using a high speed camera and tracking software the ball exits the barrel at 550 Kph.

When shot directly up it takes 19 seconds for the ball to land again, how high did the ball get?

Answer 1

Speed of the ball$(u) = 550kph=\frac{550\times 1000}{60\times 60} m/s$

$=152.77m/s$

Total time taken to land the ball at the same level = 19 sec

$h=ut-\frac{gt^2}{2}$

Now, substituting the values,

$\Rightarrow h = 152.77\times \frac{19}{2}-\frac{9.8\times 19^2}{4\times 2}$

$\Rightarrow h = (1451.315-442.225) m$

$\Rightarrow h = 1009.09m$

Hence, the maximum height reached by the ball be 1009.09m