# 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`

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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.