A ball is thrown upward from the ground. It passes a window 10 mabove the ground

Question

A ball is thrown upward from the ground. It passes a window 10 mabove the ground is seen to descend past the window 2.2safter it went by on its way up (i.e on a parabolic trajectory). ItReaches the ground 3.6 s after it was thrown. Use this informationto calculate the acceleration due to gravity, g.

This question is throwing me because I was under the impressionthat acceleration due to gravity should always be -9.8m/s2 for free falling objects close to the surface ofthe earth. Is this a trick question or am I missing a concept?

Explanation / Answer

   This problem requires using two facts    (1) Object moving under gravity takes sameto to reach ( from ground) a height h as it would take to reachground from that height.    (2)   Motion of projectilecan be studied by resloving velocity of object into vertical andhorizontal components.    Time of flightof ball   T   =   3.6   s   =   2* u * sin /g                        u* sin    =   1.8 *g             -------------(1)    here   u   =   initialspeed ofball,      =   anglewith horizon    Time taken by ball to reach the heightof window ( 10 m)   =   (3.6 - 2.2) /2   =   0.7   s    Vertical speed ofball   =   u * sin    Hence second equation of motion is    h   =   uy* t   -   (1/2) * g *t2    10   =   u *sin * 0.7   -   0.5 * g *0.72    Substitute the value of u * sin fromequation (1)    10   =   1.8 *g * 0.7   -   0.245 * g    g   =   10 /(1.8 * 0.7 -0.245)   =   9.852   m/s2    Substitute the value of u * sin fromequation (1)    10   =   1.8 *g * 0.7   -   0.245 * g    g   =   10 /(1.8 * 0.7 -0.245)   =   9.852   m/s2    

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