Question

A ball 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).

Answer 1

3.07 m/s

Step-by-step explanation:

We know that from kinematics equation

`v^(2)=u^(2)+2as` 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

`u=\sqrt {v^(2)-2gs}`

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

`u=\sqrt {6.79^(2)-(2* 9.81* 1.87)}=3.068338313 m/s\approx 3.07 m/s`