Question

the horizontal distance (d) of an object launched at an angle (\theta) with initial velocity (v_0) is given by the following formula.(d = \frac{v_0^{2}sin(2\theta)}{g})if (g = 9.8 m/s^{2}), find the horizontal distance of an object when (v_0 = 100 m/s), and (\theta=\frac{2pi}{7} rad).do not round any intermediate computations. round your answer to the nearest hundredth.(d=\text{}m)

Explanation:

Step1: Substitute given values into formula

Given $v_0 = 100$ m/s, $\theta=\frac{2\pi}{7}$ rad, and $g = 9.8$ m/s². Substitute into $D=\frac{v_0^{2}\sin(2\theta)}{g}$. First, find $2\theta$: $2\theta = 2\times\frac{2\pi}{7}=\frac{4\pi}{7}$ rad.
Then the formula becomes $D=\frac{100^{2}\sin(\frac{4\pi}{7})}{9.8}$.

Step2: Calculate $\sin(\frac{4\pi}{7})$

Using a calculator, $\sin(\frac{4\pi}{7})\approx0.9749$.

Step3: Calculate $100^{2}$

$100^{2}=10000$.

Step4: Calculate the numerator

The numerator is $10000\times0.9749 = 9749$.

Step5: Calculate $D$

$D=\frac{9749}{9.8}\approx994.7959$.
Rounding to the nearest hundredth, $D\approx994.80$.

Answer:

$994.80$ m