Question

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 60.9 m horizontally from the end of the ramp. His velocity, just before landing, is 25.2 m/s and points in a direction 31.4 ° 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

We know that in projectile motion velocity in horizontal direction remains constant, since there is no acceleration in horizontal direction, So

Vfx = Vix

Now Vfx = Vf*cos theta

Vfx = 25.2*cos 31.4 deg = 21.5 m/sec

So, Vix = 21.5 m/sec

Now given that range of motion is, R = 60.9 m

Range is given by:

R = Vix*T

T = time of motion = ?

T = R/Vix = 60.9/21.5 = 2.83 sec

Now Using 1st kinematic equation in vertical direction:

Vfy = Viy + ay*t

a = acceleration in vertical direction = -g = -9.81 m/sec^2

Vfy = -Vf*sin theta = -25.2*sin 31.4 deg = -13.1 m/sec (Here Vfy is -ve, since angle is below the horizontal)

So,

Viy = Vfy - a*t

Viy = -13.1 - (-9.81)*2.83

Viy = 14.7 m/sec = initial vertical velocity

Vix = 21.5 m/sec = initial horizontal velocity

Vi = sqrt (Vix^2 + Viy^2)

Vi = sqrt (21.5^2 + 14.7^2)

Vi = 26.04 m/sec = Magnitude of initial velocity

Direction = arctan (Viy/VIx)

Direction = arctan (14.7/21.5) = 34.4 deg above the horizontal

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