Answer:
the basketball player must leave the ground with a velocity of 4.748 m/s
Explanation:
Given that data in the question;
From the third equation of motion;
v² - u² = 2as
such that;
u² = v² - 2as
where u is the initial velocity, v is the final velocity, s is the displacement and a is acceleration
so initial velocity of the basket ball player will be;
u = √( v² - 2as )
so from the question; s is 1.15 m and a = - 9.8 m/s² { player is under negative acceleration to get to the ball } and final velocity of the player will be 0.
so we substitute
u = √( (0)² - (2 × -9.8 × 1.15 )
u = √ -( - 22.54 )
u = √ ( 22.54 )
u = 4.748 m/s
Therefore, the basketball player must leave the ground with a velocity of 4.748 m/s
