Question

A ball is thrown with an initial velocity of $100m/s$ at an angle of $30^\circ$ above the horizontal. How far from the throwing point will the ball attain its original level? Solve the problem without using formula for horizontal range.

Answer 1

In the question we were asked to find the horizontal distance using the formula for horizontal range. So, we will start with calculating the vertical distance between throwing point and original level. And using all the values we will find the horizontal distance, using the formula mentioned, i.e., $Horizontal{\text{ }}distance{\text{ }} = {\text{ }}V{\text{ }}cosx{\text{ }} \times {\text{ }}total{\text{ }}time$

Complete step by step answer:
We have been given that a ball is thrown with an initial velocity of $100m/s$ at an angle of $30^\circ$ above the horizontal.
We need to find the horizontal distance covered by the ball.
So, the given initial velocity of the ball $= {\text{ }}100m/s$
The given angle by which the ball is thrown above horizontal $=$ $30^\circ$
Since, it makes the angle above horizontal, so its vertical component $= {\text{ }}100{\text{ }}sin30^\circ$

= {\text{ }}100(\dfrac{1}{2}) \\\ = 50m/s \\\ \end{gathered} $$ The time taken by the ball to acquire maximum height will be at a point when its vertical velocity is zero. We know that, Time $ = \dfrac{{V\sin x}}{a}$ where, V $$ = $$ initial velocity a $$ = $$ acceleration due to gravity $$\left( {10{\text{ }}m/s} \right)$$ On putting the value in the above formula, we get T $ = \dfrac{{V\sin x}}{a}$ $ = \dfrac{{100(\sin 30^\circ )}}{{10}} = \dfrac{{100(\dfrac{1}{2})}}{{10}} = \dfrac{{50}}{{10}}$ $$ = {\text{ }}5{\text{ }}sec.$$ So, after $$5{\text{ }}sec$$ it will reach its maximum height. And following the symmetry, the ball will land after $$10{\text{ }}sec.$$ Now, to calculate the horizontal distance we will use the formula mentioned below. $$Horizontal{\text{ }}distance{\text{ }} = {\text{ }}V{\text{ }}cosx{\text{ }} \times {\text{ }}total{\text{ }}time$$ On putting the value in the above-mentioned formula, we get Horizontal distance $$ = {\text{ }}V{\text{ }}cosx{\text{ }} \times {\text{ }}time$$$ = 100\cos 30^\circ \times {\text{ }}10$ $\begin{gathered} = 100(\dfrac{{\sqrt 3 }}{2}) \times {\text{ }}10 \\\ = 100(\dfrac{{1.73}}{2}) \times {\text{ }}10 \\\ = 866.02 \\\ \approx 866m \\\ \end{gathered} $ Thus, the total horizontal distance from the throwing point from original level is $$866{\text{ }}m.$$ **Note:** Here, in the question we were asked about the horizontal distance of the ball. Horizontal distance is the distance between the two points measured at $$0\% $$ slope. To calculate the horizontal distance, first we need to find the vertical distance or rise and the percent of elevation between the two given points.