The maximum height H of an object launched with initial velocity v_(0) at an angle Theta is given by the following formula. H=(v_(0)^2sin^2Theta )/(2g) If g=9.8m/s^2 , find the height of an object when v_(0)=90m/s and Theta =1.83 rad. Do not round any intermediate computations. Round your answer to the nearest hundredth. H=square m

To find the height \( H \) of the object, we will use the given formula:<br /><br />\[ H = \frac{v_{0}^{2} \sin^{2} \Theta}{2g} \]<br /><br />Given:<br />- \( v_{0} = 90 \, \text{m/s} \)<br />- \( \Theta = 1.83 \, \text{rad} \)<br />- \( g = 9.8 \, \text{m/s}^2 \)<br /><br />First, we need to calculate \( \sin^2 \Theta \):<br /><br />\[ \sin \Theta = \sin(1.83) \]<br /><br />Using a calculator:<br /><br />\[ \sin(1.83) \approx 0.965925826 \]<br /><br />So,<br /><br />\[ \sin^2(1.83) = (0.965925826)^2 \approx 0.934 \]<br /><br />Now, substitute the values into the formula:<br /><br />\[ H = \frac{(90)^2 \cdot 0.934}{2 \cdot 9.8} \]<br /><br />Calculate the numerator:<br /><br />\[ (90)^2 = 8100 \]<br /><br />\[ 8100 \cdot 0.934 = 7543.4 \]<br /><br />Now, divide by the denominator:<br /><br />\[ H = \frac{7543.4}{2 \cdot 9.8} = \frac{7543.4}{19.6} \approx 384.93 \]<br /><br />Therefore, the height \( H \) of the object is approximately:<br /><br />\[ H \approx 384.93 \, \text{m} \]<br /><br />So, the the object is:<br /><br />\[ H = 384.93 \, \text{m} \]