Question

A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s. What is the maximum height the ball will reach?

Answer 1

Hope this helps a little

initial distance up = 2

initial velocity component up = 9 sin 60 = 7.79

v = 9 sin 60 - 9.8 t

when v = 0, we are there

9.8 t = 7.79

t = .795 seconds to top

h = 2 + 7.79(.795) - 4.9(.795^2)

Answer 2

Answer:

6.2m

Explanation:

Maximum height is the height reached by an object launched into space before falling under the influence of gravity. Mathematically,

Maximum height (H) = u²sin²(theta)/2g

Given initial velocity (u) = 9m/s

Angle of launch = 60°

Acceleration due to gravity g = 9.8m/s

H = 9²(sin60)²/9.8

H = 81(0.8660)²/9.8

H = 6.2m

The maximum height reached by the ball is 6.2m