A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 85 m horizontally away from and below the point of release. What is the initial velocity of the ball in the horizontal direction? Express your answer in term of m/s. Write done the number only. Keep two significant figures.

18.81m/s^2

Explanation:

Given the following :

Height of building = 0. 1 km = 100m

Horizontal distance = 85m

Using the equation :

S = 1/2gt^2

And S = 100, g = 9.8m/s^2

100 = 0.5(9.8)(t^2)

100 = 4.9(t^2)

t^2 = (100 / 4.9)

t^2 = 20.408

t = 4.5175214

t = 4.52s

Therefore, initial velocity of ball in horizontal direction;

Using the equation:

S = ut + 0.5at^2

a in horizontal direction = 0

Therefore,

S = ut

85 = u × 4.52

u = (85 / 4.52)

u = 18.805

u = 18.81m/s

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