PLEASE HELP ME SOLVE THIS!! PLZ Banpo Bridge fountain is 60 metres above the water.
The water is projected with a horizontal speed of 6m/s
-Calculate:
1. The time taken to hit the river for a single droplet of water.
2. The range of the projectile for a single droplet of water.
3. The velocity (diagonal) when it hits the water.

1.) Time = 3.5 seconds

2.) Range = 21 m

3.) Velocity = 34.8 m/s

Explanation:

Given that the

Height h = 60 m

Initial velocity U = 6 m/s

1.) The time taken to hit the river for a single droplet of water can be achieved by using second equation of motion

h = Ut + 1/2gt^2

Let assume that the water drops from rest. Therefore, U = 0

60 = 1/2 × 9.8t^2

60 = 4.9t^2

t^2 = 60/4.9

t^2 = 12.24

t = sqrt (12.24)

t = 3.5s

2.) The range of the projectile for a single droplet of water will be calculated by using the formula

R = Ut

Range R = 6 × 3.5 = 20.99

Range = 21 m

3.) The velocity (diagonal) when it hits the water.

Using third equation of motion

V^2 = U^2 + 2gH

Substitute values into the equation

V^2 = 6^2 + 2 × 9.8 × 60

V^2 = 36 + 1176

V^2 = 1212

V = sqrt (1212)

V = 34.8 m/s

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