After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 44.5 m horizontally from the end of the ramp. His velocity, just before landing, is 15.4 m/s and points in a direction 35.1 ° 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.

Question

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 44.5 m horizontally from the end of the ramp. His velocity, just before landing, is 15.4 m/s and points in a direction 35.1 ° 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.

Answer 1

To find the initial velocity of the ski jumper when he left the end of the ramp, we need to break the given information into components and use the principles of projectile motion. Let's assume the X-axis as the horizontal direction and the Y-axis as the vertical direction.

(a) To find the magnitude of the initial velocity (Vi), we can use the x-component of the velocity, Vix, since there is no acceleration in the horizontal direction (neglecting air resistance). We can find Vix using the given information:

Vix = V * cosθ, where Vix is the horizontal component of the velocity, V is the velocity magnitude, and θ is the angle below the horizontal.

Substituting the known values:
Vix = 15.4 m/s * cos(35.1°) = 12.6 m/s

So, the magnitude of the initial velocity is 12.6 m/s.

(b) To find the direction of the initial velocity, we can use the y-component of the velocity, Viy. Since we know the horizontal displacement (Δx) and the initial horizontal velocity (Vix), we can use the horizontal equation of motion to find the time of flight (t):

Δx = Vix * t

Rearranging the equation, we have:
t = Δx / Vix = 44.5 m / 12.6 m/s = 3.53 s

Now, using the time of flight (t) and the vertical equation of motion, we can find the initial vertical velocity (Viy):

Viy = g * t, where g is the acceleration due to gravity (9.8 m/s^2).

Substituting the known values:
Viy = 9.8 m/s^2 * 3.53 s = 34.59 m/s

Since the initial vertical velocity (Viy) is negative (as gravity acts downward), the direction of the initial velocity can be calculated as:

θi = arctan(Viy / Vix)

Substituting the known values:
θi = arctan(-34.59 m/s / 12.6 m/s) ≈ -71.9°

Therefore, the magnitude of his initial velocity was 12.6 m/s and it was directed at an angle of approximately -71.9° below the horizontal.

Answer 2

Use energy concepts here.

InitialPE+initialKE=final PE+final KE