Question

A ball is rolled off the edge of a swimming pool and strikes the water 2.0 m away 0.30 second later. (a) What was the initial speed of the ball? (b) What is its speed when it struck the water?

Answer 1

Constant velocity is given as follows. We can apply this to the horizontal motion because we know it to be constant as it is perpendicular to gravity. \[v=\frac{\Delta x}{\Delta t}\Rightarrow \boxed{\displaystyle v_x=\frac{\Delta x}{\Delta t}}\]The final vertical velocity is given by \[v_f = \cancel{v_i} + at \Rightarrow v_{y,f} = -g\Delta t\]The total final velocity is given by the Pythagorean theorem.\[v_f=\sqrt{v_x^2+v_{y,f}^2} \Rightarrow \boxed{\displaystyle v_f = \sqrt{\left(\frac{\Delta x}{\Delta t}\right)^2+\left(g\Delta t\right)^2}}\]