Question

a rock is thrown upward with a velocity of 12 meters per second from the top of a 25 meter high cliff, and it misses the cliff on the way back down. when will the rock be 3 meters from ground level? round your answer to two decimal places. gravity formula answer how to enter your answer (opens in new window) seconds

Explanation:

Step1: Recall the height formula for projectile motion

The height \( h(t) \) of an object in vertical motion under gravity is given by the formula \( h(t)=-4.9t^{2}+v_{0}t + h_{0} \), where \( v_{0} \) is the initial velocity, \( h_{0} \) is the initial height, and \( t \) is the time in seconds. Here, \( v_{0} = 12\space m/s \), \( h_{0}=25\space m \), and we want to find \( t \) when \( h(t) = 3\space m \). So we set up the equation:
\( 3=-4.9t^{2}+12t + 25 \)

Step2: Rearrange the equation to standard quadratic form

Subtract 3 from both sides to get:
\( - 4.9t^{2}+12t+25 - 3=0 \)
\( -4.9t^{2}+12t + 22 = 0 \)
Multiply both sides by - 1 to make the coefficient of \( t^{2} \) positive:
\( 4.9t^{2}-12t - 22=0 \)

Step3: Use the quadratic formula

The quadratic formula for a quadratic equation \( ax^{2}+bx + c = 0 \) is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). Here, \( a = 4.9 \), \( b=- 12 \), \( c=-22 \).
First, calculate the discriminant \( D=b^{2}-4ac \):
\( D=(-12)^{2}-4\times4.9\times(-22) \)
\( D = 144+431.2 \)
\( D=575.2 \)
Then, find \( t \):
\( t=\frac{12\pm\sqrt{575.2}}{2\times4.9} \)
We have two solutions:
\( t_{1}=\frac{12+\sqrt{575.2}}{9.8}\approx\frac{12 + 23.98}{9.8}\approx\frac{35.98}{9.8}\approx3.67 \)
\( t_{2}=\frac{12-\sqrt{575.2}}{9.8}\approx\frac{12-23.98}{9.8}\approx\frac{- 11.98}{9.8}\approx - 1.22 \)
Since time cannot be negative, we discard \( t_{2} \).

Answer:

\( 3.67 \)