Find Velocity of 2nd Ball Thrown at 25 Degrees - 6.32 m/s

[BunDa4Th]

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

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.

 

[berkeman]

I don't think you need any squares for this problem. The vertical component of the velocity of the 2nd ball needs to equal the vertical component of the first ball's velocity. What is the vertical component of the 2nd ball's velocity?

 

[BunDa4Th]

Okay, I was finally able to figure this out when I took my test today. Thanks for the input. I made a mistake of thinking that how far must the ball go to reach the same height distance of the first which was a big mistake I did.