Question #2567f

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Answer 1 · 2015-06-27T15:37:50

24.5m

To get the height the ball reaches above the plaza we can use:

$s = u t + \frac{1}{2} g t^{2}$ $s=ut+1/2"g"t^(2)$

It will take 4/2 = 2s to reach its zenith.

$g = - 9.8 m/s$ $"g"=-9.8"m/s"$

$s = 19.6 \times 2 + \frac{1}{2} g 2^{2}$ $s=19.6xx2+1/2"g"2^(2)$

$= 39.2 - 19.6 = 19.6 m$ $=39.2-19.6=19.6"m"$

We can get the total distance the ball falls using:

$S_{tot} = \frac{1}{2} g t^{2}$ $S_("tot")=1/2"g"t^(2)$

This time $g = + 9.8 m/s$ $"g"=+9.8"m/s"$

and the total time = $2 + 1 = 3 s$ $2+1 = 3s$

$S_{tot} = \frac{1}{2} g 9 = \frac{9 g}{2}$ $S_("tot")=1/2"g"9=(9"g")/2$

If the height of the tower is $h$ $h$ then:

$19.6 + h = \frac{9 g}{2}$ $19.6+h=(9"g")/(2)$

$h = 44.1 - 19.6 = 24.5 m$ $h=44.1-19.6=24.5"m"$