Connect with experts and get insightful answers to your questions on IDNLearn.com. Discover reliable and timely information on any topic from our network of experienced professionals.
Sagot :
For a) You know that when the ball reaches maximum hight, it has a velocity of 0m/s. You would use the equation [tex] v^{2} = v_{0} ^{2}+2ax [/tex]... V is the velocity at the apex and that is 0m/s, [tex]v_{0}[/tex] is the initial velocity which is 15.0m/s, 'a' is the acceration and in this case it is gravity which will be -9.8m/s^2 because we decided that up will be positive, and "x" is the distance between your hand and the apex. So you will have 0= 15^2+2(-9.8)x and you should get 11.48m.
For b) you would use the equation x=[tex] x_{0}+v_{0}t+ \frac{1}{2}at^{2} [/tex]. In this case x would 11.48m, [tex] x_{0} [/tex] is 0m, and t is time.. so you would have 11.48=15t+1/2(-9.8)t^2 and get t= 1.5s ... this is just the time it took to go to its maximum hight so you multiply it by 2 because it takes the same time to get up as it does to come down. You should of gotten 3 seconds. I hope this helped
For b) you would use the equation x=[tex] x_{0}+v_{0}t+ \frac{1}{2}at^{2} [/tex]. In this case x would 11.48m, [tex] x_{0} [/tex] is 0m, and t is time.. so you would have 11.48=15t+1/2(-9.8)t^2 and get t= 1.5s ... this is just the time it took to go to its maximum hight so you multiply it by 2 because it takes the same time to get up as it does to come down. You should of gotten 3 seconds. I hope this helped
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.
