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 =
-5 because the denominator can’t equal zero
Answer:
-h - 6
Step-by-step explanation:
(6h + 2) + (-7h - 8) <== remove parenthesis
6h + 2 - 7h - 8 <== combine like terms
6h - 7h + 2 - 8
-h - 6
Hope this helps!
Answer:
39.6x +26.4
Step-by-step explanation:
The area of a rectangle is given by
A = l*w
A = 13.2 * (3x+2)
Distribute
39.6x +26.4
Make a rectangle that is 10 squares by 10 squares and color in 11 squares.
