A diver leaves the end of a 4.0-m-high diving board and strikes the water 1.3 s later, 3.0 m beyond the end of the board. Considering the diver as a particle, determine:

Answer 1

(a) the horizontal component of the initial velocity

(b) the vertical component of the initial velocity

(c) the magnitude of the initial velocity

We will solve this problem using the kinematic equations.

(a) To find the horizontal component of the initial velocity, we will use the horizontal displacement formula:
x = x0 + vx * t

where x is the horizontal displacement, x0 is the initial horizontal position (which can be considered as 0), vx is the horizontal component of the initial velocity, and t is the time elapsed.

We are given x = 3.0 m and t = 1.3 s. Plugging these values into the equation, we get:

3.0 m = 0 + vx * 1.3 s

Now, we can solve for vx:

vx = 3.0 m / 1.3 s ≈ 2.31 m/s

Thus, the horizontal component of the initial velocity is approximately 2.31 m/s.

(b) To find the vertical component of the initial velocity, we will use the vertical displacement formula:
y = y0 + vy * t - 0.5 * g * t^2

where y is the vertical displacement, y0 is the initial vertical position (which can be considered as 0), vy is the vertical component of the initial velocity, g is the acceleration due to gravity (9.81 m/s^2), and t is the time elapsed.

We are given y = -4.0 m (negative because it is downward) and t = 1.3 s. Plugging these values into the equation, we get:

-4.0 m = 0 + vy * 1.3 s - 0.5 * 9.81 m/s^2 * (1.3 s)^2

Solving for vy, we get:

vy ≈ -0.87 m/s

Thus, the vertical component of the initial velocity is approximately -0.87 m/s (negative because it is downward).

(c) To find the magnitude of the initial velocity, we will use the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)

Plugging in the values we found for vx and vy, we get:

v = sqrt((2.31 m/s)^2 + (-0.87 m/s)^2) ≈ 2.46 m/s

Thus, the magnitude of the initial velocity is approximately 2.46 m/s.