Final answer:
The velocity at which the long jumper landed can be computed by separating the jump into horizontal and vertical components, using the initial take-off angle and the horizontal distance. The horizontal and vertical velocities are combined to find the magnitude of the total landing velocity, and the arctangent of their ratio gives the direction.
Step-by-step explanation:
To determine the velocity at which the long jumper landed, we need to consider the projectile motion of the jump. There are two components of velocity to consider: the horizontal (vx) and the vertical (vy) at the point of landing.
First, the horizontal velocity (vx) can be found by dividing the total horizontal distance by the time of flight (t). The equation of horizontal motion is:
vx = d / t, where d is the horizontal distance.
The vertical velocity (vy) of the jumper when he lands will be the same magnitude but opposite in direction to the vertical velocity at take-off due to symmetry in projectile motion, assuming no air resistance. The vertical velocity at take-off can be calculated using the initial jump angle (Ө) and the initial velocity (v0).
Using the initial angle of 30°, the vertical component of the initial velocity at take-off (v0y) is:
v0y = v0 * sin(Ө).
Since the vertical motion is subject only to acceleration due to gravity (g), the final vertical velocity at landing (vy) will be v0y (but in the opposite direction).
The total landing velocity (v) is then found by combining these two components using the Pythagorean theorem:
v = √(vx^2 + vy^2).
The direction of the landing velocity is given by the angle made with the horizontal, found using the arctangent of the ratio of vy over vx:
θ = arctan(vy / vx).
These calculations would provide the magnitude and direction of the landing velocity.