A ball of mass 0.152 kg is dropped from a height 2.27 m above the ground. The acceleration of gravity is 9.8 m/s 2 . Neglecting air resistance, determine the speed of the ball when it is at a height 0.903 m above the ground. Answer in units of m/s.

8.44 m/s.

Explanation:

Change in Potential Energy = Mass x Acceleration From Gravity x H2 - H1

Kinetic Energy = 1/2 (mass) x [(v2)^2 - v1^2]

g * h = 1/2 * v^2

(9.8) x (2.27) = 1/2 * (v)^2

v^2 = 2[(9.8) x (2.27)]

v = 6.67 m/s

g * delta h = 1/2 * delta v^2

(9.8) x (2.27 - 0.903) = 1/2 * [(v2)^2 - (6.67)^2]

v2^2 = 71.2821

v2 = 8.44 m/s

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer