Q45P - University Physics with Modern Physics

Question

A sly 1.5kg monkey and a jungle veterinarian with a blow-gun loaded with a tranquilizer dart are 25m above the ground in trees 70 m apart. Just as the veterinarian shoots horizontally at the monkey, the monkey drops from the tree in a vain attempt to escape being hit. What must the minimum muzzle velocity of the dart be for the dart to hit the monkey before the monkey reaches the ground?

Step-by-Step Solution

Answer

The minimum muzzle velocity will be $\mathrm{v}=31\mathrm{m}/\mathrm{s}$.

Step 1: Newton’s law

The velocity contains two components, one is the horizontal component and another one is vertical. 

 

According to the newton’s laws of motion,

 

$\mathbf{s}\mathbf{=}\mathbf{ut}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{at}}^{\mathbf{2}}$                                                                                                              (1)                                    

 

Where, s, u, t, a, are displacement, initial velocity, time and acceleration.

Step 2: Given data

• Weight of monkey = 1.5 kg 

 

• Displacement = 25 m

 

• Distance = 70 m   

Step 3: Minimum muzzle velocity

From the newton’s law of motion equation (1), we can calculate the time as:

$$\begin{gathered} 25 \mathrm{m}=0\times \mathrm{t}+\frac{1}{2}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times {\mathrm{t}}^2 \\ \mathrm{t}=2.259 \mathrm{s} \end{gathered}$$ 

 

 Now, for the dart hits the monkey the muzzle velocity should be,

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{x}}{\mathrm{t}} \\ \mathrm{v}=\frac{70 \mathrm{m}}{2.259 \mathrm{s}} \\ \mathrm{v}=31 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Both the objects will hit the ground at the same time. Minimum muzzle velocity of the dart be for the dart to hit the monkey before the monkey reaches the ground is v=31m/s.