Aball thrown straight up into the air is found to be moving at 6.79 m/s after falling 1.87 m below its release point. find the ball's initial speed (in m/s).

3.07 m/s

Explanation:

We know that from kinematics equation

and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

explanation:

could i give you a exaplo that could you?

i think it's 35 meters