Q48P - University Physics with Modern Physics

Question

A movie stuntwoman drops from a helicopter that is 30.0m above the ground and moving with a constant velocity whose components are 10.0m/s upward and 15.0m/s horizontal and toward the south. Ignore air resistance. (a) Where on the ground (relative to the position of the helicopter when she drops) should the stuntwoman have placed foam mats to break her fall? (b) Draw $\mathbf{x}\mathbf{-}\mathbf{t}\mathbf{,}\mathbf{y}\mathbf{-}\mathbf{t}\mathbf{,}{\mathbf{v}}_{\mathbf{x}}\mathbf{-}\mathbf{t}\mathbf{,}{\mathbf{andv}}_{\mathbf{y}}\mathbf{-}\mathbf{t}$ graphs of her motion.

Step-by-Step Solution

Answer

(a) The mats should be placed at a distance 55.5 m from the south.

(b) The graph is shown in step 4.

Step 1: 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,

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ 

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

Step 2: Given data

Distance of the helicopter from the ground = 30.0 m

 

Vertical component of the velocity ${\mathrm{v}}_{\mathrm{y}}=10.0\mathrm{m}/\mathrm{s}$

 

Horizontal component of the velocity ${\mathrm{v}}_{\mathrm{x}}=15.0\mathrm{m}/\mathrm{s}$ 

Step 3: Stuntwoman’s foam mats

(a)

 

For the vertical motion from the law of motion of Newton, 

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Substitute 30.0m for s, 10 m/s for u and $9.8m/s^2$ for a in the above equation.

$$\begin{gathered} 30.0=10\times \mathrm{t}+\frac{1}{2}\times 9.8\times {\mathrm{t}}^2 \\ \mathrm{t}=3.70\mathrm{s} \end{gathered}$$ 

 

In the time 3.70s the horizontal distance travelled by her,

$s=ut+\frac{1}{2}a{t}^2$ 

  

Substitute 15.0m/s for u, 3.70s for t and $9.8m/s^2$ for g in the above equation.

$$\begin{gathered} \mathrm{s}=15.0\times 3.70+\frac{1}{2}\times 9.8\times {\left(3.70\mathrm{s}\right)}^2 \\ \mathrm{s}=55.5\mathrm{m} \end{gathered}$$ 

 

Therefore, the mats should be placed at a distance 55.5m from the south.

Step 4 : Graph of x - t , y - t , v x - t , andv y - t

(b)

 

x-t Graph 

 

y-t Graph

$y_x-t$Graph

$v_y-t$Graph