Question

444 41111111111111 14) (290 points) A student, standing on the ground tosses a baseball directly up into the air with a speed of 3.4 m/s. The student's hand 2.5 m above the ground when the ball is released. At the same instant, another student, standing on a raised platform 1.2 m above the ground, tosses a second baseball directly up into the air with an unknown speed. The student's hand is 2.5 m above the platform when the ball is released. The two baseballs require the same amount of time to reach their maximum height. 40-0- Y.-25 x-xo = Vot+ 043 yu=1.2mX Y = 3.7m " üyo? 15-0 =3.4t+ (-9.8362 yra.sm as 3.44 +1 -4.9142 y-2.5mx y=? Vo -3.4 mls/-34 V- height 400 18- 1-49 VO = ? Cu=0 -0.9 -4.9t² 9- 9. &msz, -4.94 log2-0.ts 9 - 9.8mse t- 0.43 x Ve=v141062) (a) How long in seconds) do the baseballs take to reach their maximum heights? - V A N Answer: 0.43 x - S USE V = vtat 0.35 35 (units) lo .-254% x O level? (b) What is the first baseball's maximum height above ground level? Answer: 3:5X m 3.09 (units) (c) What is the second baseball's maximum height above ground level? Answer: 13m 5 4.3 (units) (d) What was the launch speed of the second baseball? Answer: 0.56 m2 Y 3.4 (units) -10 ix-yo vilt 15-12 12= V² + 2016 51. X -10

please help me set this up correctly
i am confused as to which formulas to use. pictures and step by step work will help. thank you.

Answer 1

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for student as 1 → U= 3.4 m/s. So H = 0.5897 m. U2 = eg - total heigh from group = 2.500.5897 = 3.09 m. – 3.09 m Answer of ® V L V = Votat 0 = 3.4 - gt t = 3.4 ' 0 = vo'-gt Vó - (9.8) (0-3469) or vo= 3.4 m/s = 0.3469 s So time = 0.35 sec also ② → Vo = 3.4 m/s H = voh = 0.5897m So Max. height: 0.5897 e 1-2 +2.5 = 413mno ©

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