Aball is thrown directly downward with an initial speed of 7.65 m/s froma height of 29.0 m. after what time interval does it strike the ground?

1.77 s

Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:

xf= xo+vo*t+(g*t^2)/2  we can consider the origin on the top so the xo=0 and xf=29 m; then

(g*t^2)/2+vo*t-xf=0  vo is the initail velocity, vo=7.65 m/s

then by solving the quadratric equation in t

t=1.77 s

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explanation: