A ball is thrown vertically upward with a speed of 18m/sec from the edge of 85m tall building. a. What is the maximum height in which the ball reached from the ground?
b. How much later it took the ball to reach the ground?
c. What is the final velocity of the ball as it hits the ground?

Explanation:

Using the formula to calculate the maximum height

H = u²/2g

u is the initial velocity = 18m/s

g is the acceleration  due to gravity  = 9.81

H = 18²/2(9.81)

H = 324/19.62

H = 16.51m

The maximum height in which the ball reached from the ground = 85m + 16.51m = 101.5m

b) Time of fight = 2u/g

T = 2(18)/g

T = 36/9.81

T = 3.67s

It took the ball 3.67s later to reach the ground

3) To get the final velocity of the ball as it hits the ground, we need to calculate the horizontal component of the velocity,

Ux = Ucosθ

ux = 18cos 90 (angle of launch is 90 since the ball is thrown vertically upwards)

Ux = 18(0)

Ux = 0m/s

Hence the final velocity of the ball as it hits the ground is 0m/s

conservation of mass

mass before something happens = mass afterwards