Answer:
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
