A diver achieves a horizontal velocity of 3.76 m/s from a diving platform located 6m above the water. How far from the edge will the diver be when she hits the water?
Xo = 3.76 m/s.
0.5g*t^2 = 6 m.
4.9t^2 = 6
t^2 = 1.22
Tf = 1.11 s.= Fall time.
Dx = Xo*Tf = 3.76m/s * 1.11s. = 4.16 m.
To find the distance from the edge that the diver will be when she hits the water, we can use the equations of motion.
The horizontal velocity of the diver remains constant throughout the fall, so we can use the equation:
d = v * t
where:
d is the horizontal distance from the edge
v is the horizontal velocity (3.76 m/s)
t is the time it takes for the diver to hit the water
To find the time, we can use the equation of motion for vertical motion:
h = ut + (1/2) * g * t^2
where:
h is the height of the platform (6m)
u is the initial vertical velocity (0 m/s)
g is the acceleration due to gravity (-9.8 m/s^2)
t is the time
Rearranging the equation, we get:
t^2 - (2h/g)t - (2h/g) = 0
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac))/2a
where:
a = 1
b = - (2h/g)
c = - (2h/g)
Substituting the values, we get:
t = (-(2h/g) ± √((2h/g)^2 - 4*(1)*(-(2h/g)))) / 2*(1)
Simplifying further, we get:
t = (√(4h^2/g^2 + 8h/g)) / 2
Substituting the values of h = 6m and g = 9.8 m/s^2, we can calculate t.
Using the value of t, we can then calculate the horizontal distance d using the equation:
d = v * t
Substituting the values of v = 3.76 m/s and t, we can calculate the distance.
By solving these equations, we can find the distance from the edge that the diver will be when she hits the water.