A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.15 m above the floor in an attempt to get the ball

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

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

see earlier answer