Question

A telemetry system is used to quantify kinematic
values of a ski jumper immediately before she
leaves the ramp. According to the system r =
500ft, ̇r = −105ft/sec, ̈r = −10 ft/sec2
, θ = 25◦
,

̇θ = 0.07rad/sec, ̈θ = 0.06rad/sec2

. Determine
(a) the velocity of the skier immediately before
she leaves the jump, (b) the acceleration of the
skier at this instant, (c) the distance of the jump
d neglecting lift and air resistance.

Answer 1

velocity of the skier V = Vrey + Voco () 20 V= rer + reco 25° Vro velocity of skier in 25° radial direction Vo: of Skier in velocity transverse direction r = -105 ft/s 7 = 500 ft Ò = 0.07 rad/s Substituting the values in eqr We get V - (-105) er + (500) (0.07) eo V = -105 er t 35 eo 3) V- > Magnitude of velocity V- (-105)2 + (35) VE 110.67 ft/s eo 135 35 sin 65 er * Resolution of polar Components into Tectangular Components → 35 (0565 65 105 cos 25 i 25 105 + 105 Sin 25 Cs Scanned with CamScanner

V= (-105 cos 25 i – las sin 25 j) + (-35 cos 65 i + 35 sin 65j) (-105 cos 25 - 35 cos 65) i + (-los sin 25 +35 Sin 65) (-109.95) i t (-12.65); velocity in rectangular Components! Angie between the components of tan a Vy V- Va tan a = -12.65 -109.95 = 6.563° 6] acceleration vectors a- ar er + ao co a - (ö - roz) er t crö + 2řoeo 9 = 0.06 vad/se Ò= 0.07 rad/s rs 500 ft ☺ = -lo ft/s2 ☺ = -105 ft/s we substituting the values e get [(-10) - 50000.07)*]er + [500 (0.06) + 2 (-105) (0:07)], a - -12.45 er + 15.3 co of * Magnitude acceleration (-12.45)2 + (15.3)2 a = 19.725 ft/se a: CS Scanned with CamScanner

* Rewriting the components rectangular by resolving a: acceleration ego in terms of recto polar components (-12-45 cos 25 i 12.45 Sin 25 j) + (-15-3 Cos 65 i + 15.3 Sin 65j) - 17. 7495 it 8.6049 j a: the between Angle Components of tan B ay da tan ß 8.6049 -17.7495 - 25.99 c] Jumping fix of rectangular coordinate from the end of ramp is a projectile motion. at the end of ramp. origin is towards the pt d and they downward fix initial position ongin. Initial position is tve 2 measurement of skier at ToO DO Vac co .: Do = 109.95 ft/s +7- To + tot as of 109.95t X = 109.95 t Yo-0 Initial position vga yo CS Scanned with CamScanner

jo = 12.965 ft/s 9 = a + got + / t* 9: o + J2: 65 t +. - (32-2) + 9= 12.65 t + 16.1t2 * Cos 30° d X = dcos 30° Sin 30 - y-lo d 6 d sin 30° 4-10- Sin 30° multiplying ear @ by a sin 30º = d Cos 30° sio 30° Cos 30" by Multiplying eqr Cos 30ºCy-10) -> substract eq? © by d sin 30° COS 309 from eq? 5 d sin 30° COS 30° - d Cos 30°sinzo X sin 30 COS 30° 30° Cy-10) a sin 30° – COS 30" (7-10) - 0 Substituting value of 14 y. we get 109.95 t sin 30° - COS 30° (12.65 t + 16.172 -10) -O Solving above -13.943 t2 + 44.0217 t +8.66020 above eqa, we t=-0.1858 eget t-3.343 CS Scanned with CamScanner

considering positive value for t, 109.95 (3.343) 367. 569 Ft X from eq^ a & d COS 30° 367. 569- d cos 30' d=424.43 ft CS Scanned with CamScanner