Answer:
6.48m/s
Step-by-step explanation:
Given:
Distance between gun and a back =2.1 m
height 1 = 1.6m
height 2 = 0.93m
Final velocity=0 m/sec
RTC=Initial velocity.
SOLUTION
FROM THE THIRD EQUATION OF A UNIFORMLY ACCELERATED MOTION,
V^2=U^2 + 2as
0 m/s^2 = U^2 + 2×10m/s^2×2.1m
0 m/s = U^2 + 42s^2
-U^2 = 42 s^2 - 0m/s
∫-U^2= ∫42m/s
-U = 6.48m/s
-1 -1
U= 6.48m/s
★The velocity of water when it lefts the gun is 6.48 m/s