Question

4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The

horizontal range of the ball is R, and the ball reaches a maximum height R
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 θi

Answer 1

Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time
`t`, the horizontal position
`x` and vertical position
`y` of the ball are given respectively by

`x = v_i \cos(\theta_i) t`

`y = v_i \sin(\theta_i) t - \frac g2 t^2`

and the horizontal velocity
`v_x` and vertical velocity
`v_y` are

`v_x = v_i \cos(\theta_i)`

`v_y = v_i \sin(\theta_i) - gt`

The ball reaches its maximum height with
`v_y=0`. At this point, the ball has zero vertical velocity. This happens when

`v_i \sin(\theta_i) - gt = 0 \implies t = \frac{v_i \sin(\theta_i)}g`

which means

`y = \frac R6 = v_i \sin(\theta_i) * \frac{v_i \sin(\theta_i)}g - \frac g2 \left(\frac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \frac{6{v_i}^2 \sin^2(\theta_i)}g - \frac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \frac{3{v_i}^2 \sin^2(\theta_i)}g`

At the same time, the ball will have traveled half its horizontal range, so

`x = \frac R2 = v_i \cos(\theta_i) * \frac{v_i \sin(\theta_i)}g \\\\ \implies R = \frac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g`

Solve for
`v_i` and
`\theta_i` :

`\frac{3{v_i}^2 \sin^2(\theta_i)}g = \frac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0`

Since
`0^\circ<\theta_i<90^\circ`, we cannot have
`\sin(\theta_i)=0`, so we're left with (e)

`3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \frac23 \\\\ \implies \boxed{\theta_i = \tan^(-1)\left(\frac23\right) \approx 33.7^\circ}`

Now,

`\cos\left(\tan^(-1)\left(\frac23\right)\right) = \frac3{√(13)}`

`\sin\left(\tan^(-1)\left(\frac23\right)\right) = \frac2{√(13)}`

so it follows that (d)

`R = \frac{2{v_i}^2 *\frac3{√(13)} * \frac2{√(13)}}g \\\\ \implies {v_i}^2 = (13Rg)/(12) \\\\ \implies \boxed{v_i = \sqrt{(13Rg)/(12)}}`

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

`v_y = \sqrt{(13Rg)/(12)} \sin\left(\tan^(-1)\left(\frac23\right)\right) \\\\ \implies v_y = \sqrt{(13Rg)/(12)} * \frac2{√(13)} \\\\ \implies \boxed{v_y = \sqrt{\frac{Rg}3}}`

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

`v_x = \sqrt{(13Rg)/(12)} * \frac3{√(13)} \\\\ \implies v_x = (√(3Rg))/(2)`

Then with
`v_y=0`, the ball's speed
`v` is

`v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \frac{√(3Rg)}2}`

Finally, in the work leading up to part (e), we showed the time to maximum height is

`t = \frac{v_i \sin(\theta_i)}g`

but this is just half the total time the ball spends in the air. The total airtime is then

`2t = \frac{2 * \sqrt{(13Rg)/(12)} * \frac2{√(13)}}g \\\\ \implies 2t = 2\sqrt{\frac R{3g}}`

and the ball is in the air over the interval (a)

`\boxed{0 < t < 2\sqrt{\frac R{3g}}}`