Question

the horizontal distance d of an object launched at an angle θ with initial velocity ( v_0 ) is given by the following formula.
( d = \frac{v_0^2 sin(2\theta)}{g} )
if ( g = 9.8 , \text{m/s}^2 ), find the horizontal distance of an object when ( v_0 = 80 , \text{m/s} ), and ( \theta = 0.34 , \text{rad} ).
do not round any intermediate computations. round your answer to the nearest hundredth.
( d = square , \text{m} )

Explanation:

Step1: Calculate \( v_0^2 \)

Given \( v_0 = 80 \, \text{m/s} \), so \( v_0^2 = 80^2 = 6400 \).

Step2: Calculate \( 2\theta \)

Given \( \theta = 0.34 \, \text{rad} \), so \( 2\theta = 2\times0.34 = 0.68 \, \text{rad} \).

Step3: Calculate \( \sin(2\theta) \)

Using a calculator, \( \sin(0.68) \approx 0.6234 \) (keep more decimals for accuracy).

Step4: Calculate the numerator \( v_0^2\sin(2\theta) \)

Multiply the results from Step1 and Step3: \( 6400\times0.6234 = 3989.76 \).

Step5: Calculate \( D \) using the formula \( D = \frac{v_0^2\sin(2\theta)}{g} \)

Given \( g = 9.8 \, \text{m/s}^2 \), so \( D = \frac{3989.76}{9.8} \approx 407.12 \).

Answer:

\( 407.12 \)