QuestionA ball is dropped from a 59 ft building.
(a) Time to fall half the distance? $t=$
(b) Time to reach the ground? $t=$ sec

Studdy Solution

STEP 1

Assumptions1. The initial height from which the object is dropped, $h_{0}$ , is59 ft. . The height $h$ after time $t$ seconds is given by the equation $h=-16 t^{}+h_{0}$ .
3. We need to find the time it takes for the object to fall half the distance and the total distance to the ground level.

STEP 2

First, we need to find the time it takes for the object to fall half the distance to the ground level. This means the height $h$ is half of the initial height $h_{0}$ , i.e., $h = h_{0}/2$ . We can substitute this into the equation for $h$ and solve for $t$ .
$-16 t^{2}+h_{0} = h_{0}/2$

STEP 3

Substitute the given value for $h_{0}$ into the equation.
$-16 t^{2}+59 =59/2$

STEP 4

olve the equation for $t^{2}$ .
$t^{2} = \frac{59/2 -59}{-16}$

STEP 5

Calculate the value of $t^{2}$ .
$t^{2} = \frac{59/2 -59}{-16} =1.84375$

STEP 6

Since $t$ is time and cannot be negative, we take the positive square root of $t^{2}$ to find $t$ .
$t = \sqrt{1.84375}$

STEP 7

Calculate the value of $t$ .
$t = \sqrt{1.84375} =1.358$ So, it will take approximately1.358 seconds for the object to fall half the distance to the ground level.

STEP 8

Next, we need to find the time it takes for the object to fall to the ground level. This means the height $h$ is0. We can substitute this into the equation for $h$ and solve for $t$ .
$-16 t^{2}+h_{0} =0$

STEP 9

Substitute the given value for $h_{}$ into the equation.
$-16 t^{2}+59 =$

STEP 10

olve the equation for $t^{2}$ .
$t^{2} = \frac{59}{16}$

STEP 11

Calculate the value of $t^{}$ .
$t^{} = \frac{59}{16} =3.6875$

STEP 12

Again, since $t$ is time and cannot be negative, we take the positive square root of $t^{2}$ to find $t$ .
$t = \sqrt{.6875}$

STEP 13

Calculate the value of $t$ .
$t = \sqrt{3.6875} =.921$ So, it will take approximately.921 seconds for the object to fall to the ground level.

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