# Question 00f8c

Nov 25, 2017

The horizontal travel before landing would be farther when the angle is ${35}^{\circ}$.

#### Explanation:

Your first answer showed that the horizontal component of velocity would be less with the angle at ${35}^{\circ}$. But the vertical component of velocity would be greater. Because of the greater vertical velocity, it will spend more time in the air. That allows the horizontal velocity to carry the ball farther. (It turns out that ${45}^{\circ}$ gives the maximum range.)

We can get the time from a vertical motion analysis:
$s = u \cdot t + a \cdot {t}^{2} / 2$
where s = 0 (because it ends up on the ground),
$u = 13.9 \frac{m}{s} \cdot \sin {35}^{\circ} \left(\text{initial vertical component of velocity}\right)$, and
a = -9.8 m/s^2.

Plugging in the data
$0 = 13.9 \frac{m}{s} \cdot \sin {35}^{\circ} \cdot t + \frac{\left(- 9.8 \frac{m}{s} ^ 2\right) \cdot {t}^{2}}{2}$

Dividing both sides by t
$0 = 13.9 \frac{m}{s} \cdot \sin {35}^{\circ} + \frac{\left(- 9.8 \frac{m}{s} ^ 2\right) \cdot t}{2}$

Solving for t
$\frac{9.8 \frac{m}{s} ^ 2 \cdot t}{2} = 13.9 \frac{m}{s} \cdot \sin {35}^{\circ}$

$t = \frac{2 \cdot 13.9 \frac{m}{s} \cdot \sin {35}^{\circ}}{9.8 \frac{m}{s} ^ 2} = 1.62 s$

Now a horizontal motion analysis:

The horizontal component of the initial velocity is $13.9 \frac{m}{s} \cdot \cos {35}^{\circ}$ and it does not change until the ball hits the ground. So the horizontal distance it travels is given by multiplying that velocity by the time in the air.
$s = 13.9 \frac{m}{s} \cdot \cos {35}^{\circ} \cdot 1.62 s = 18.5 m$

I hope this helps,
Steve