Question

ball is projected horizontally from a height of 30m above the ground with a speed of 20m/s. calculate the horizontal distance travelled by the ball when it hits the ground g = 10m/s2]​

Answer 1

The range is                      

                                    $\Large\displaystyle\text{ \begin{gathered} \text{Range } = R= v_x\sqrt{\frac{2S_{y_0}}{g}}\\ \\R= 20\sqrt{6} \approx 48.98\text{ m}\\ \\\end{gathered} }$

We know that in cinematic problems the velocity in different directions is independent, so on the y-axis, or the y-velocity we have the gravity acceleration changing the velocity, although on the x-axis, or the x-velocity we don't have any acceleration, so the velocity is constant, so, we can use the basic cinematic equations and adapt them for our problem:

                                     $\Large\displaystyle\text{ \begin{gathered}S_x = S_{x_0} + v_x t\\ \\v_y = v_{y_0} - gt \\ \\S_y = S_{y_0} + v_{y_0}t - \frac{gt^2}{2}\end{gathered} }$

We can simplify it, we know that all the initial position on the x-axis is zero, but the initial position on the y-axis is 30, we also know that the initial velocity on y-axis is zero, so

                                           $\Large\displaystyle\text{ \begin{gathered}S_x = v_x t\\ \\v_y =- gt \\ \\S_y = S_{y_0} - \frac{gt^2}{2}\end{gathered} }$

I won't put numbers know, let's do all the manipulation first.

When the ball hit the ground we know that his position is 0 on the y-axis, so

                                           $\Large\displaystyle\text{ \begin{gathered}0 = S_{y_0} - \frac{gt^2}{2}\\ \\S_{y_0} = \frac{gt^2}{2}\\ \\t = \sqrt{\frac{2S_{y_0}}{g}}\end{gathered} }$

Now we know when the ball hits the ground, so we can use this time on the position equation on the x-axis and find out the distance the ball travelled

                                          $\Large\displaystyle\text{ \begin{gathered}S_x = v_x t\\ \\S_x = v_x\sqrt{\frac{2S_{y_0}}{g}}\end{gathered} }$

That's all! we can just put the values know.

                                $\Large\displaystyle\text{ \begin{gathered} \text{Range } = R= v_x\sqrt{\frac{2S_{y_0}}{g}}\\ \\R= 20\sqrt{\frac{2\cdot 30}{10}}\\ \\R= 20\sqrt{6} \approx 48.98\text{ m}\\ \\\end{gathered} }$

Hope you liked it

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