QuestionA rock is thrown up at 12 m/s from a 42 m cliff. When is it 12 m above ground? Round to two decimal places.

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the rock is12 m/s. The initial height from which the rock is thrown is42 m3. The acceleration due to gravity is -9.8 m/s² (negative because it's acting downwards)
4. The rock misses the cliff on the way back down5. We need to find the time when the rock is12 m from the ground level

STEP 2

We can use the equation of motion to solve this problem. The equation is $h = h0 + v0t - \frac{1}{2}gt^2$ where- $h$ is the height at time $t$ - $h0$ is the initial height- $v0$ is the initial velocity- $g$ is the acceleration due to gravity- $t$ is the time

STEP 3

We want to find the time when the rock is12 m from the ground level. So, we set $h$ to $12$ and plug in the given values for $h0$ , $v0$ , and $g$ .
$12 =42 +12t - \frac{1}{2}(-9.8)t^2$

STEP 4

implify the equation.
$0 = -\frac{1}{2}(9.8)t^2 +12t + (42 -12)$

STEP 5

This is a quadratic equation in the form of $0 = at^2 + bt + c$ . We can solve it for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

STEP 6

Plug in the values for $a$ , $b$ , and $c$ into the quadratic formula to find the values of $t$ .
$t = \frac{-12 \pm \sqrt{12^2 -4(-\frac{1}{2}(9.8))(42 -12)}}{2(-\frac{1}{2}(9.8))}$

STEP 7

olve the equation to find the values of $t$ .
$t \approx1.49,5.71$ The rock is12 m from the ground level at approximately1.49 seconds and5.71 seconds after it was thrown.

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