Question

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.)

Answer 1

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

$\Delta t = 4.1 - 0.5$

$\Delta t = 3.6 s$

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

$t = \frac{\Delta t }{2}$

$t = 1.8 s$

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

$T = 1.8 + 0.5 = 2.3 s$

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

$v_f - v_i = at$

$0 - v_i = -9.8 * 2.3$

$v_i = 22.54 m/s$

also the speed after 0.5 s

$v_f - 22.54 = -9.8* 0.5$

$v_f = 17.64 m/s$

so the height of the pole is given as

$h = \frac{v_f + v_i}{2}*t$

$h = \frac{22.54 + 17.64}{2}*0.5 = 10 m$

so height of the pole is 10 m