Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m above the floor at an angle of 60° above the horizontal, what must the initial speed be (in m/s) if it were to go through the basket?

The answer to your question is    vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

              hmax = vo²sinФ/ 2g

Solve for vo²

              vo² = 2ghmax / sinФ

Substitution

              vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

              vo² = 19.62(1.3) / 0.866

              vo² = 25.51 / 0.866

              vo² = 29.45

Result

              vo = 5.43 m/s

it was given as a lesson and guide for prayer.

higher potential

explanation:

because it is in a high speed so it must move into a of high potential