Hello sopheary dara mao,
I would orient the vertical $x$-axis of motion such that the origin coincides with the initial position of the stone, and model the motion of the stone with the IVP (where $x$ is measured in feet and time $t$ is measured in seconds):
$$x''(t)=-g$$ where $$x'(0)=v_0$$ and $$x(0)=x_0$$
Integrating once, we find:
$$x'(t)=v(t)=-gt+C$$
Using the initial conditions, we find:
$$v_0=C$$ hence:
$$v(t)=-gt+v_0$$
Integrating again, we find:
$$x(t)=-\frac{g}{2}t^2+v_0t+C$$
Using the initial conditions, we find:
$$x_0=C$$ hence:
$$x(t)=-\frac{g}{2}t^2+v_0t+x_0$$
Now, using the given data and orientation of our coordinate system, i.e:
$$g=32,\,v_0=14,\,x_0=0$$ we have:
$$x(t)=-16t^2+14t$$
$$v(t)=-32t+14$$
Now we are ready to answer the questions:
A.) $$x(2)=-16(2)^2+14(2)=-36$$
This means after 2 seconds the stone is 36 feet below its initial position, or 764 ft above the ground.
B.) To find at what time the stone hits the ground, we may set:
$$x(t)=-800$$
$$-16t^2+14t=-800$$
$$16t^2-14t-800=0$$
$$2\left(8t^2-7t-400 \right)=0$$
Applying the quadratic formula, and taking the positive root, we find:
$$t=\frac{7+\sqrt{12849}}{16}\approx7.5221$$
C.) $$v\left(\frac{7+\sqrt{12849}}{16} \right)=-32\left(\frac{7+\sqrt{12849}}{16} \right)+14=-2\sqrt{12849}$$
The negative sign indicates the stone is moving in a downward direction when it hits the ground. This is in agreement with the speed when using energy considerations, i.e., equating initial energy (gravitational potential and kinetic) to final energy (kinetic) and solving for the final velocity.
To sopheary dara mao and any other guests viewing this topic, I invite and encourage you to post other kinematics questions in either our http://www.mathhelpboards.com/f10/ forum if the calculus is to be used or our http://www.mathhelpboards.com/f22/ forum if energy considerations, or other pre-calculus techniques are to be used.
Best Regards,
Mark.
