Q47P - University Physics with Modern Physics

Question

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. For practice, a pilot drops a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path 90.0 m above the ground and has a speed of$\mathbf{64}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{(}\mathbf{143}\mathbf{ }\mathbf{mi}\mathbf{/}\mathbf{h}\mathbf{)}$, at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

Step-by-Step Solution

Answer

The canister should be released from the horizontal distance s =274m from the target.

Step 1: Given data

Distance of the plane from the ground = 90.0m 

 

Speed = 64.0m/s

Step 2: Introduction

If a particle is thrown in projectile motion, then the velocity of the particle will contain two components, first one is horizontal and other one is vertical.

 

According to the Newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$ 

Here s is the displacement,u is the initial velocity,a is the acceleration and t is the time.

Step 3: Horizontal distance

The projectile motion of the canister is shown below,

The initial speed of the canister is zero.

 

From newton’s law of motion, for the vertical motion

$s=ut+\frac{1}{2}a{t}^2$           

Substitute 90.0 m for s, 0m/s for u,$9.8m/s^2$ for a in the above equation.

$$\begin{gathered} 90.0=0\times t+\frac{1}{2}\times 9.8\times t^2 \\ t=4.286s \end{gathered}$$ 

           

Now from the horizontal component of the motion the distance cover by canister in the 4.286s,

 

From the newton’s law of motion,

$s=ut+\frac{1}{2}a{t}^2$           

 

Substitute 64.0ms for u, 4.286 s for t, $9.8m/s^2$ for a in the above equation.

$$\begin{gathered} s=64.0\times 4.286+\frac{1}{2}\times 9.8\times 4.286 \\ s=274m \end{gathered}$$ 

 

 

Therefore, the pilot will release the canister at a distance of 274m from the target.