Question

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 36.0 m horizontally from the end of the ramp. His velocity, just before landing, is 19.0 m/s and points in a direction 33.2 ° below the horizontal. Neglecting air resistance and any lift that he experiences while airborne, find (a) the magnitude and (b) the direction of his initial velocity when he left the end of the ramp. (a) Number Units

Answer 1

To find the magnitude and direction of the initial velocity when the ski jumper left the end of the ramp, we can use trigonometry. The horizontal displacement, or range, is given as 36.0 m and the velocity just before landing is 19.0 m/s at an angle of 33.2° below the horizontal.

Step-by-step explanation:

To find the magnitude and direction of the initial velocity when the ski jumper left the end of the ramp, we can use trigonometry. The horizontal displacement, or range, is given as 36.0 m and the velocity just before landing is 19.0 m/s at an angle of 33.2° below the horizontal.

(a) To find the magnitude of the initial velocity, we can use the range formula: range = initial velocity * time. Since the vertical component of the velocity does not affect the range, we can use the horizontal component of the initial velocity: 36.0 m = initial velocity * cos(33.2°). Solving for the magnitude of the initial velocity gives us: initial velocity = 36.0 m / cos(33.2°).

(b) To find the direction of the initial velocity, we can use the tangent of the angle: tan(angle) = vertical component / horizontal component. Rearranging the equation gives angle = arctan(vertical component / horizontal component). Substituting the given values, we get: angle = arctan(19.0 m/s * sin(33.2°) / 19.0 m/s * cos(33.2°)).