To determine the height of a flagpole, abby throws a ball straight up and times it. she sees that the ball goes by the top of the pole after 0.50 s and then reaches the top of the pole again after a total elapsed time of 4.1 s. how high is the pole above the point where the ball was launched? (you can ignore air resistance.)

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aristocles aristocles · Answer 1 · 2018-09-14T03:37:08+02:00

Initially the ball pass the pole after 0.5 s

and then after 4.1 s again it will reach the top of the pole

So time taken by the ball to cover the distance above the pole is

[tex]\Delta t = 4.1 - 0.5[/tex]

[tex]\Delta t = 3.6 s[/tex]

now the time it will take to reach the highest point from the top of the pole is given as

[tex]t = \frac{\Delta t }{2}[/tex]

[tex]t = 1.8 s[/tex]

now the total time to reach the top from the bottom is given as

[tex]T = 1.8 + 0.5 = 2.3 s[/tex]

so the initial speed by which it will throw is given as

[tex]v_f - v_i = at[/tex]

[tex]0 - v_i = -9.8 * 2.3 [/tex]

[tex]v_i = 22.54 m/s[/tex]

also the speed after 0.5 s

[tex]v_f - 22.54 = -9.8* 0.5[/tex]

[tex]v_f = 17.64 m/s[/tex]

so the height of the pole is given as

[tex]h = \frac{v_f + v_i}{2}*t[/tex]

[tex]h = \frac{22.54 + 17.64}{2}*0.5 = 10 m[/tex]

so height of the pole is 10 m