A ball is thrown upward with an initial velocity of 9.8 m/s. How high does it reach before it starts descending? Choose only one from the three formulas: 1. Vf = Vi + gt 2. dy = Viyt + 1/2gdyt^2 3. Vfy = Viy^2 + 2gdy

To solve, apply the third formula.
v_(fy)^2 = v_(iy)^2+2gd_y
Take note that when the ball reaches the maximum height, its velocity is zero. So plugging in the values 
v_(iy)=9.8 m/s
v_(fy) = 0
g=-9.8 m/s^2
the formula becomes
0^2= 9.8^2 + 2(-9.8)d_y
0=96.04 - 19.6d_y
19.6d_y = 96.04
d_y=96.04/19.6
d_y=4.9
Therefore, the maximum height of the ball is 4.9 meters.
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