A discus is thrown from a height of 3 feet with an initial velocity of 55 ft/s at an angle of 44° with the horizontal. How long
will it take for the discus to reach the ground? I started off with 55cos(44)=40 then I made an equation 40t-16(t)=2.5 seconds? I feel like I did this wrong because I didn't add 3 feet anywhere
or do I add 3 in to 55cos(44) giving me an answer of 2.6
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions. Yes, the 3 feet are important. <span>You can consider the datum (y=0) at 3' from the ground, and the ground is therefore at -3 feet. </span>
<span>The vertical initial velocity, vy, is u*sin(θ), so it gives </span> <span>vy=55sin(44°)=38.21 approx. </span>
<span>The (vertical) distance travelled is given by: </span> <span>S=vy*t-(1/2)gt² </span> <span>where </span> <span>S=-3 (ground) </span> <span>vy=55sin(44°), and </span> <span>g=32.2 ft/s² </span> <span>Solve for t. </span> <span>I get -0.08 and 2.45 s.</span>
The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ... No; The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ANSWER: No; 9.
