10 or 25 and a lot more pretty much anything that ends with 0 or 5
3.01 m/s
This is a simple projectile calculation. What we want is a vertical velocity such that the time the droplet spends going up and going back down to the surface exactly matches the time the droplet takes to travel horizontally 0.800 meters. The time the droplet spends in the air will is:
V*sqrt(3)/2 ; Vertical velocity.
(V*sqrt(3)/2)/9.8 ; Time until droplet reaches maximum height
(V*sqrt(3))/9.8 ; Double that time for droplet to fall back to the surface.
The droplet's horizontal velocity will be:
V/2.
So the total distance the droplet travels will be:
d = (V*sqrt(3))/9.8 * V/2
d = V^2*sqrt(3)/19.6
Let's substitute the desired distance and solve for V
d = V^2*sqrt(3)/19.6
0.8 = V^2*sqrt(3)/19.6
15.68 = V^2*sqrt(3)
15.68/sqrt(3) = V^2
15.68/1.732050808 = V^2
3.008795809 = V
So after rounding to 3 significant figures, the archerfish needs to spit the water at a velocity of 3.01 m/s
Let's verify that answer.
Vertical velocity: 3.01 * sin(60) = 3.01 * 0.866025404 = 2.606736465
Time of flight = 2.606736465 * 2 / 9.8 = 0.531987034 seconds.
Horizontal velocity: 3.01 * cos(60) = 3.01 * 0.5 =
Answer:
of wht?....................
You multiply 5 x 3 Bc you multiplied the left side by 3. Hope this helps
