At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Ok, so since we know the angle of "takeoff" and the total distance and time we can find the velocity, now the issue with this velocity is that it is the vertical velocity, not the velocity in the y-direction. However, if you've taken geometry you probably know about SOH-CAH-TOA, and you were probably taught to do this problem using triangles. Since you have the angle, and the hypotenuse (the vertical velocity you solved for) then you can find the x and y parts that make up the vertical velocity. Now I'm a bit confused as to wether you are solving for the distance it goes up or the total distance it goes overall.
Answer:
Explanation:
Maximum height = U²sin²angle/2g
Where u is velocity
Angle is in degree
g is acceleration due to gravity
Time of flight = 2usinangle/g
Range = U²sine2 x angle/g
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.