-4/8 i think that is right not sure tho so you might wanna wait for ohter answer
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:
15 ft
Step-by-step explanation:
Hi, the illustration for the problem is a right triangle with:
hypotenuse (C)= the length of the ladder = L
horizontal side(A) = distance from bottom of the ladder to the building = L - 6
vertical side(B) = distance from the top of the ladder to the bottom of the building = L - 3
So, we can use Pythagoras formula:
A2 +B2= C2
(L – 6 )² + (L-3)² = L²
L²-12L+36+L²-6L+9 =L²
L2 -18L+45 =0
APPLYING QUADRATIC FORMULA WE OBTAIN:
L =15 OR L=3
If L=3
Vertical side = L-3 = 0 (Length can´t be 0)
So L=15
Y=.25x is the only resonable answer
