A ball is thrown straight upward and returns to the thrower's hand after 2.50 s in the air. A second ball is thrown at an angle of 25.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically? m/s(adsbygoogle = window.adsbygoogle || []).push({});

okay, I am not sure what I am doing wrong here:

to find deltaY it would be deltay = -1/2(g)(t)^2 = deltay -1/2(-9.8)(2.50)^2

deltay = 30.63

now i know that deltaX = 30.63 because that is the same distance the first ball is.

V_0y = Vosin25 and V_0x = Vocos25

when setting up the eqaution i should get 30.63 = (V_0^2)/(9.8) x sin2(25)

30.63/sin2(25) = Vo^2/9.8 which then become Vo^2 = 39.98 Vo = 6.32

which is incorrect. What mistake or what steps should I be following.

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# Homework Help: Velocity question

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