Q11E - University Physics with Modern Physics

Question

Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while Milada jumps horizontally with an initial speed of 95.0 cm/s. How far from the base of the cliff will Milada hit the ground? Ignore air resistance

Step-by-Step Solution

Answer

The distance of Milada from the base of cliff is 2.57 s.

Step 1: Identification of given data:

The given data can be listed below,

The time taken by chirpy to reach ground is ${\mathrm{t}}_{\mathrm{c}}=2.70\hspace{0.33em}\mathrm{s}$.

The initial speed of Milada is, ${\mathrm{v}}_{\mathrm{i},\mathrm{m}}=95\hspace{0.33em}\frac{\mathrm{cm}}{\mathrm{s}}$. 

Step 2: Concept/Significance of the displacement

An object's total change in position is referred to as displacement, it is a vector variable that measures how far out of place an item is.

Step 3: Determination the distance of the base of the cliff will Milada hit the ground:

The distance of Milada from the base of the cliff is given by,

$\mathrm{x}-{\mathrm{x}}_0={\mathrm{v}}_0\mathrm{xt}+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2$ 

Here, ${\mathrm{v}}_{0\mathrm{x}}$ is the initial velocity of Milada, t is the time to reach the ground, and ${\mathrm{a}}_{\mathrm{x}}$ is the horizontal component of acceleration whose value is zero.

 

Substitute all the values in the above expression.

$$\begin{gathered} \mathrm{x}-{\mathrm{x}}_0=\left(0.950\mathrm{m}/\mathrm{s}\right)\left(2.70 \mathrm{s}\right)+\frac{1}{2}\left(0\right){\left(2.70 \mathrm{s}\right)}^2 \\ =2.565+0 \\ =2.57 \mathrm{sa} \end{gathered}$$ 

 

Thus, the distance of Milada from the base of the cliff is 2.57 s.