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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
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