Astudent drops a ball from a window 3.5 m above the sidewalk. how fast is the ball moving when it hits the ground?

To answer this you must know that the acceleration caused by gravity upon the ball, or any object, is -9.8 m/s^2
there's two ways to solve this
first, use the position equation
x=(1/2)at^2+Vot+Xo
Xo, the initial position, is 3.5 m
x(t), the final position, is 0 m (the ground)
Vo, the initial velocity, is 0 m/s, since you just drop the ball
and a is -9.8
so
0=(1/2)(-9.8)t^2+3.5
-3.5=-4.9t^2
t^2=0.71
t=0.845
so it takes the ball 0.845 s to hit the ground
now, using the velocity equation, v=at+Vo,
v=(-9.8)(0.845)+0=-8.28 m/s
therefore, the speed of the ball is 8.28 m/s when it hits the ground

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