A high diver leaves the end of a 10m high diving board andstrikes the water 2.6

Question

A high diver leaves the end of a 10m high diving board andstrikes the water 2.6 s later, 6.0 m beyond the end of theboard. a) calculate the initial velocity of the diver and express itin unit vector form. b) calculate the maximum height reached by thediver. c.)calculate the velocity of the diver just before she entersthe water and express it in unit vector form. if you could do both x and y components and show all work.Thank you!!! A high diver leaves the end of a 10m high diving board andstrikes the water 2.6 s later, 6.0 m beyond the end of theboard. a) calculate the initial velocity of the diver and express itin unit vector form. b) calculate the maximum height reached by thediver. c.)calculate the velocity of the diver just before she entersthe water and express it in unit vector form. if you could do both x and y components and show all work.Thank you!!!

Explanation / Answer

   a.   Range    R   =   Horizontalcomponent of velocity * time       6m   =   ux *2.6   =>   initial velocity alongXaxis   ux   =   6/2.6   =   2.31   m/s       Let the vertical componentby uy. Total time in air (t = 2.6 s) could be brokeninto two parts, t1 during which the diver goes up andreaches board level ( 10 m above water surface) and t2during which the diver falls to water ( height 10 m)       t   =   t1   +   t2             t1   =   2* uy /g   =>   uy   =   g* t1 / 2   =   4.9 *t1            h   =   uy* t2   +   (1/2) * g *t22   =   4.9 *t1 * t2   +  (1/2)* 9.8 * t22            h   =   uy* t2   +   (1/2) * g *t22   =   4.9 *t1 * t2   +  (1/2)* 9.8 * t22             10   =  4.9* t1 *t2   +   4.9 *t22             2.04   =   t2* ( t1 + t2)       Substitute   t2   =   t- t1   =   2.6 -t1       andsolve          t1   =   1.81   s,   t2   =   0.79   s          uy   =   9.8* 1.81 /2   =   8.87   ms          Assumingupwards direction as positive          initialvelocity      u   =   ux* i^   +   uy * j^             u   =   2.31i^   +   8.87 j^    i^   =   unitvector along X horizontal direction,   j^   =   unit vector alongvertical.   b.    Height reached aboverboard is given by      H   =   uy2/ 2 * g   =   8.872/ 2 *9.8   =   4.01   m          Totalmaximum height above water level   Hmax   =   h+ H   =   10 +4.01   =   14.01   m    c.   Horizontal component ofvelocity remains the same throughoutflight   i.e.      vx   =   ux          Verticalcomponent of final velocity can be calculated using third equationof motion.          vy2   =   u2   +   2* g * Hmax          u   =   initialvelocity when fall begins at maximumheight   =   0          =>   vy    =   2 * 9.8 *14.01   =   16.57   m/s          Finalvelocity   v   =   vxi^   +   vyj^   =   2.31 i^  - 16.57 j^       vertical component is take-ve as it is downwards now.          u   =   initialvelocity when fall begins at maximumheight   =   0          =>   vy    =   2 * 9.8 *14.01   =   16.57   m/s          Finalvelocity   v   =   vxi^   +   vyj^   =   2.31 i^  - 16.57 j^       vertical component is take-ve as it is downwards now.

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