A baseball is thrown downward 8

A baseball is thrown downward from a 50-ft tower with an initial speed of 18 ft/s. Determine the speed at which it hits the ground and the time of travel.

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Solution:

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We will use the following formula to solve this question to figure out the speed at which it hits the ground.

$v_2^2=v_1^2+2a_c(s_2-s_1)$

(Where $v_2$ is final velocity, $v_1$ is initial velocity, $a_c$ is constant acceleration, $s_2$ is initial displacement, and $s_1$ is final displacement)

From the question, we know the following values:

$v_1=18$ ft/s

$a_c=32.2 \,\text{ft/s}^2$

$s_2=50$ ft

$s_1=0$ ft

Let us substitute these values into our equation:

$v_2^2=v_1^2+2a_c(s_2-s_1)$

$v_2^2=18^2+2(32.2)(50-0)$

$v_2^2=3544$

(Take the square root of both sides)

$v_2=59.5$ ft/s

To figure out the time of travel, we will use the following equation.

$v_2=v_1+a_ct$

(Where $v_2$ is final velocity, $v_1$ is initial velocity, $a$ is constant acceleration and $t$ is time)

Again, substituting the values we know gives us:

$v_2=v_1+a_ct$

$59.5=18+32.2(t)$

(Solve for t)

$t=1.29$ s

$v_2=59.5$ ft/s
$t=1.29$ s