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

Answer:

3.07 m/s

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

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