Answer :
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
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
