You would like to find the height of flagpole on campus. Youthrow a rock straigh

Question

You would like to find the height of flagpole on campus. Youthrow a rock straight up and find that the rock reaches the top ofthe flagpole on the way up after 0.80 s, and passes the top ofthe flagpole again on the way down 1.3 s after it wastossed. What is the height of the flagpole What is the initial speed of the rock I do not know how to set up the equation. Does it not matterthe height of the person tossing the rock?
You would like to find the height of flagpole on campus. Youthrow a rock straight up and find that the rock reaches the top ofthe flagpole on the way up after 0.80 s, and passes the top ofthe flagpole again on the way down 1.3 s after it wastossed. What is the height of the flagpole What is the initial speed of the rock I do not know how to set up the equation. Does it not matterthe height of the person tossing the rock?

Explanation / Answer

    Time for which the rock is above theflagpole   =   1.3 -0.8   =   0.5   s    Time for rock to rech highest point ( fromtop offlagpole)   t1    =  0.5/2   =   0.25   s    Total time of upwardsflight   t   =   timeto reach top offlagpole   +   t1    =   0.8+0.25   =   1.05   s    First equation of motionis   vf   =   vi   +   g* t    0   =   vi   +   (- 9.8) * 1.05   =>   initialspeed of rock   vi   =      10.29   m/s    ( g is taken as -ve as rock is thownupwards)    Second equation of moion can be used to findout height of flagpole    h   =   vi* t'   +   (1/2) * g *t'2    t'   =   0.8   s    h   =   10.29* 0.80   + 0.5 * ( -9.8) * 0.802    h   =   8.232   -   3.136   =   5.096   m    Time for rock to rech highest point ( fromtop offlagpole)   t1    =  0.5/2   =   0.25   s    Total time of upwardsflight   t   =   timeto reach top offlagpole   +   t1    =   0.8+0.25   =   1.05   s    First equation of motionis   vf   =   vi   +   g* t    0   =   vi   +   (- 9.8) * 1.05   =>   initialspeed of rock   vi   =      10.29   m/s    ( g is taken as -ve as rock is thownupwards)    Second equation of moion can be used to findout height of flagpole    h   =   vi* t'   +   (1/2) * g *t'2    t'   =   0.8   s    h   =   10.29* 0.80   + 0.5 * ( -9.8) * 0.802    h   =   8.232   -   3.136   =   5.096   m    h   =   8.232   -   3.136   =   5.096   m

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