A baseball player runs 27.4 meters from thebatter’s box to first base, overruns first base by
3.0 meters, and then returns to first base.
Compared to the total distance traveled by the
player, the magnitude of the player’s total
displacement from the batter’s box is
(1) 3.0 m shorter (3) 3.0 m longer
(2) 6.0 m shorter (4) 6.0 m longer

Explanation:

To find the takeoff speed of the long jumper, we can utilize the physics principles of projectile motion. Given that the long jumper leaves the ground at a 30-degree angle and travels a distance of 8.50 m, we need to find the initial velocity (takeoff speed) of the jumper.

In projectile motion, we can break down the motion into horizontal and vertical components. The horizontal component remains constant, while the vertical component is affected by gravity.

To solve for the takeoff speed, we can focus on the vertical component of motion. The equation that relates the vertical displacement, initial velocity, launch angle, and acceleration due to gravity is as follows:

Δy = v₀y t + (1/2) g * t²,

where:

- Δy is the vertical displacement (8.50 m),

- v₀y is the vertical component of initial velocity (takeoff speed),

- t is the total time of flight, and

- g is the acceleration due to gravity (approximately 9.8 m/s²).

Since the vertical displacement at the peak of the jump is zero (the jumper is at the highest point), we can rewrite the equation as:

0 = v₀y * t + (1/2) g t².

However, we can derive a relation between the time of flight t and the initial velocity v₀y by using the launch angle θ. The time of flight is given by:

t = (2 v₀y sin(θ)) / g.

Substituting this expression for t in the above equation, we have:

0 = v₀y [(2 v₀y sin(θ)) / g] + (1/2) g [(2 v₀y sin(θ)) / g]².

Now, we can solve for v₀y:

0 = v₀y² (2 sin(θ) + sin²(θ)) / g.

Rearranging and isolating v₀y, we get:

v₀y = √[(g Δy) / (2 * sin(θ) + sin²(θ))].

With the given values:

Δy = 8.50 m,

θ = 30 degrees,

g ≈ 9.8 m/s²,

we can substitute these values into the formula:

v₀y = √[(9