A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B ,3.00 m ,higher than A, with speed$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{v}$Calculate (a) the speed v and (b) the maximum height reached by the stone above point B.

1. Speed of the stone at point A is v
2. Speed of the stone at point B is$\frac{1}{2}\mathrm{v}$
3. Point B is higher than point A by$3.00\mathrm{m}$

The problem deals with the kinematic equation of motion. Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The final velocity in the kinematic equation is given by,

${{\mathrm{v}}_{\mathrm{f}}}^2={{\mathrm{v}}_0}^2+2\mathrm{ay}\dots ..........................................................................\left(\mathrm{i}\right)$

Where, is the initial velocity, a is acceleration, and y is the vertical displacement.

Let’s write the equation (i) at point A.

${\mathrm{v}}^2={{\mathrm{v}}_0}^2-2\mathrm{g}\cdot \mathrm{y}........................................................................\left(\mathrm{i}\right)$

Here y is the height at point A and $-\mathrm{g}$ is the gravitational acceleration. Since we assume the upward direction is positive and acceleration is in the downward direction, the sign of g is negative.

Similarly, write the equation (i) for the point B

${\left(\frac{1}{2}\mathrm{v}\right)}^2={{\mathrm{v}}_0}^2-2\mathrm{g}(\mathrm{y}+3)\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(\mathrm{iii}\right)$

Now,add equations (ii) and (iii)

$$\begin{gathered} {\mathrm{v}}^2-\frac{1}{4}{\mathrm{v}}^2={{\mathrm{v}}_0}^2-2\mathrm{g}.\mathrm{y}-{{\mathrm{v}}_0}^2+2\mathrm{g}\left(\mathrm{y}+3\right) \\ \frac{3}{4}{\mathrm{v}}^2=2\mathrm{g}\left(3\right) \\ {\mathrm{v}}^2=8\mathrm{g} \\ \mathrm{v}=8.85\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the speed would be $\mathrm{v}=8.85\mathrm{m}/\mathrm{s}$

Now, lets assume that the initial velocity is $8.85\mathrm{m}/\mathrm{s}$. When the stone reach the maximum height, it’s velocity will be zero. Using equation (i), we can write that

$$\begin{gathered} 0={\left(8.85\right)}^2-2\mathrm{g}.∆\mathrm{y} \\ ∆\mathrm{y}=\frac{{\left(8.85\mathrm{m}/\mathrm{s}\right)}^2}{2\mathrm{g}} \\ =4\mathrm{m} \end{gathered}$$

Therefore, the stone can reach a maximum height of 4 m from point A.

Since point B is at 3 m height from point A, the maximum height above point is, $4.00\hspace{0.33em}\mathrm{m}-3.00\mathrm{m}=1.00\hspace{0.33em}\mathrm{m}$.

Therefore, the maximum height reached by the stone above point B is 1.00 m.