Question

A tiger leaps horizontally from a 7.5-m-high rock with a speed of$\mathbf{3}\mathbf{m}\mathbf{/}\mathbf{s}$. How far from the base of the rock will she land?

Short answer

The tiger lands at a distance of $3.708\mathrm{m}$from the base of the rock.

Step-by-step solution

Step 1. Significance of the motion of the tiger

The analysis of the motion of the tiger can be performed with the help of the equations of motion.

The tiger leaps in the horizontal direction. The horizontal distance moved by the tiger can be evaluated with the help of the tiger's speed and the time taken by her.

Step 2. Identification of the given data

The given data can be listed below as:

• The height of the rock is $y_0=7.5\mathrm{m}$.
• The speed of the tiger is $v_x=3\mathrm{m}/\mathrm{s}$.
• The acceleration acting on the tiger in the downward direction is $a=g=-9.81\mathrm{m}/{\mathrm{s}}^2$.

Step 3. Determination of the time taken by the tiger and the distance moved by her 

The horizontal speed of the tiger can be expressed as:

$\begin{array}{c}{v}_x=\frac{d_x}{t}\\ d_x=v_{x}t\dots \left(i\right)\end{array}$

Here, $d_x$is the horizontal distance covered by the tiger from the base of the rock.

The vertical distance covered by the tiger can be expressed as:

$\begin{array}{c}y=y_0+v_{0}t+\frac{1}{2}a{t}^2\\ =y_0+v_{0}t+\frac{1}{2}g{t}^2\end{array}$

Here, g is the acceleration due to gravity whose value is $9.81\mathrm{m}/{\mathrm{s}}^2$; $y_0$ is the vertical distance of the rock from the ground, which is equal to the initial distance moved by the tiger in the y-direction; y is the final distance of the tiger, which is taken as 0 m as the tiger comes to rest; and $v_0$is the initial speed of the tiger in the vertical direction whose value is 0 m/s.

Substitute the values as $\left(-9.81\mathrm{m}/{\mathrm{s}}^2\right)$for g, 0 m for y, 0 m/s for$v_0$, and $7.5\mathrm{m}$for $y_0$in the above equation.

$\begin{array}{c}0\mathrm{m}=7.5\mathrm{m}+\left(0\mathrm{m}/\mathrm{s}\right)\times t+\frac{1}{2}\times \left(-9.81\mathrm{m}/{\mathrm{s}}^2\right)t^2\\ \frac{1}{2}\times \left(9.81\mathrm{m}/{\mathrm{s}}^2\right)t^2=7.5\mathrm{m}\\ t=\sqrt{1.529}\mathrm{s}\\ =1.236\mathrm{s}\end{array}$

Substitute the values as $3\mathrm{m}/\mathrm{s}$ for $v_x$, and $1.236\mathrm{s}$ for t in equation (I).

$\begin{array}{c}{d}_x=3\mathrm{m}/\mathrm{s}\times 1.236\mathrm{s}\\ =3.708\mathrm{m}\end{array}$

Thus, she lands at a distance of $3.708\mathrm{m}$from the base of the rock.