Question

A ball is thrown with an initial speed${\mathit{v}}_{\mathbf{i}}$at an angle${\mathit{\theta }}_{\mathbf{i}}$with the horizontal. The horizontal range of the ball is R, and the ball reaches a maximum height$\mathit{R}\mathbf{/}\mathbf{6}$. In terms of R and g, find (a) the time interval during which the ball is in motion, (b) the ball's speed at the peak of its path, (c) the initial vertical component of its velocity, (d) its initial speed, and (e) the angle${\mathit{\theta }}_{\mathbf{c}}$. (f) Suppose the ball is thrown at the same initial speed found in (d) but at the angle appropriate for reaching the greatest height that it can. Find this height. (g) Suppose the ball is thrown at the same initial speed but at the angle for greatest possible range. Find this maximum horizontal range.

Short answer

(a) The time interval during which the ball is in motion is $2\sqrt{\frac{R}{3g}}$.

(b) The balls speed at the beak of the path is $\frac{\sqrt{3gR}}{2}$.

(c) The initial vertical component of the velocity is $\frac{\sqrt{gR}}{3}$.

(d) The initial velocity of the ball is $\sqrt{\frac{13gR}{12}}$.

(e) The angle${\theta }_c$ is $33.7°$.

(f) The maximum height reached by the object is $\frac{13}{24}R$.

(g) The maximum horizontal range of the ball is $\frac{13}{12}R$.

Step-by-step solution

Understanding the concept

The expression for the maximum height reached in case of the projectile motion is given by,

$H=\frac{v_i^2{\sin }^2{\theta }_i}{2g}$

The expression for the total time of flight in case of projectile motion is given by,

$T=\frac{2v_i\sin {\theta }_i}{g}$

Here H is the maximum height, $v_i$is the initial velocity, ${\theta }_i$ is the initial angle from the horizontal, g is the acceleration due to gravity, T is total time of flight.

(a) time duration in which ball is motion 

Consider vertical component of initial velocity is $v_{yi}$, and the maximum height reached by the ball is $h=R/6$. Then from the equation of motion we have

$\begin{array}{rcl}{v}_{yi}^2& =& 2gh\\ & =& 2g\frac{R}{6}\\ & =& \frac{gR}{3}\end{array}$

$\therefore v_{yi}=\sqrt{\frac{gR}{3}}$

Now, if the time taken by ball to reach the maximum height is t, then we have

$\begin{array}{rcl}{v}_{yi}-gt& =& 0\\ gt& =& v_{yi}\\ t& =& \frac{v_{yi}}{g}\end{array}$

Now, substituting the value of$v_{yi}$ in the above equation we have

$$\begin{gathered} t=\frac{1}{g}\sqrt{\frac{gR}{3}} \\ t=\sqrt{\frac{R}{3g}} \end{gathered}$$

Now, this is the time required by the ball to travel in one direction. Hence, the total time duration for which the ball was in motion will be twice the above calculated time. Hence, the time duration in which the ball was in motion is

$$\begin{gathered} T=2T \\ T=2\sqrt{\frac{R}{3g}} \end{gathered}$$

Therefore, the time interval during which the ball is in motion is $2\sqrt{\frac{R}{3g}}$.

(b) Horizontal component of ball’s velocity 

At the peak of its path, the vertical component of the ball’s velocity is zero. Hence, at the peak of its path, the balls speed is the horizontal component of the velocity. Now the horizontal range of the ball is and time taken by the ball to reach the distance is T, hence, the horizontal component of the ball's velocity is given by

$\begin{array}{rcl}{v}_x& =& \frac{R}{T}\\ & =& \frac{R}{2\sqrt{\frac{R}{3g}}}\end{array}$

$\therefore v_x=\frac{\sqrt{3gR}}{2}$

Therefore the balls speed at the beak of the path is $\frac{\sqrt{3gR}}{2}$.

(c) Initial vertical component of the velocity

As calculated in the equation (1), the initial vertical component of the velocity is

$v_{yi}=\sqrt{\frac{gR}{3}}$

Therefore the initial vertical component of the velocity is $\sqrt{\frac{gR}{3}}$.

(d) Initial speed of the ball

The initial speed of the ball is

$\begin{array}{rcl}v& =& \sqrt{v_{yi}^2+v_x^2}\\ v& =& \sqrt{\left(\frac{gR}{3}\right)+\left(\frac{3gR}{4}\right)}\\ v& =& \sqrt{\frac{13gR}{12}}\end{array}$

Therefore the initial velocity of the ball is $\sqrt{\frac{13gR}{12}}$.

(e) Initial angle θi

The initial angle ${\theta }_i$is given by

$\begin{array}{rcl}\tan {\theta }_i& =& \frac{v_{yi}}{v_x}\\ & =& \frac{\sqrt{gR/3}}{\sqrt{3gR/4}}\\ & =& \frac{2}{3}{\theta }_i\end{array}$

$\therefore {\tan }^{-1}\left(\frac{2}{3}\right)=33.7°$

Therefore the angle ${\theta }_c$is $33.7°$.

(f) Maximum height reached by the ball

The ball will reach greatest possible height, the ball was thrown directly upward, that is if there is not horizontal component of the velocity. So, in this case the maximum height reached by the ball is given by

$\begin{array}{rcl}h& =& \frac{v^2}{2g}\\ & =& \frac{{\left(\sqrt{\frac{13gR}{12}}\right)}^2}{2g}\end{array}$

$\therefore h=\frac{13}{24}R$

Therefore the maximum height reached by the object is $\frac{13}{24}R$.

(g) range when the projectile is thrown

Projectile travels greatest possible range when the projectile is thrown at an $\theta =45°$. In this case the range is given by

$$\begin{gathered} R_{\max }=\frac{v^2}{g} \\ {R}_{\max }=\frac{\left(\sqrt{{\left(\frac{13gR}{12}\right)}^2}\right)}{g} \end{gathered}$$

Therefore the maximum horizontal range of the ball is $\frac{13}{12}R$.