What is the horizontal distance travelled by a ball if it is thrown with an initial velocity of 20ms^-1 at 30^0?

 To find the horizontal distance traveled by the ball, we can use the following projectile motion equations.

The horizontal distance (${�}_{�}$) can be calculated using the formula:

${�}_{�}={�}_{�}\cdot �\cdot \cos (�)$

where:

• ${�}_{�}$ is the initial velocity,
• $�$ is the time of flight,
• $�$ is the angle of projection.

Firstly, we need to calculate the time of flight ($�$) using the vertical motion equation:

$�=\frac{2\cdot {�}_{�}\cdot \sin (�)}{�}$

where:

• $�$ is the acceleration due to gravity (approximately 9.8 m/s²).

Given ${�}_{�}=20�\mathrm{/}�$ and $�=30^{\circ }$, let's calculate $�$ first:

$�=\frac{2\cdot 20\cdot \sin (30)}{9.8}$

$�\approx \frac{2\cdot 20\cdot 0.5}{9.8}$

$�\approx \frac{20}{9.8}$

$�\approx 2.04�$

Now that we have the time of flight, we can use it to find the horizontal distance ${�}_{�}$:

${�}_{�}=20\cdot 2.04\cdot \cos (30)$

${�}_{�}\approx 20\cdot 2.04\cdot \frac{\sqrt{3}}{2}$