Q66P - University Physics with Modern Physics

Question

Henrietta is jogging on the sidewalk at 3.05m/s on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is 38.0m above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally 9.0s after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?

Step-by-Step Solution

Answer

(a) u =12.9m/s

 

(b) s =35.9m

Step 1: Newton’s law

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

 

According to the newton’s laws of motion,

 

$s=ut+\frac{1}{2}a{t}^2$

 

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

 

For the vertical and horizontal motion, the velocity component will be respectively.

Step 2: Given

Henrietta jogging speed = 3.05m/s

 

Window vertical distance from the street = 38.0m

                                                                                           

Passed time = 9.0s

Step 3: (a) Initial Speed of throwing the bagels

The initial velocity is zero, so from the vertical motion equation,

$$\begin{gathered} \mathrm{s}=\frac{1}{2}{\mathrm{gt}}^2 \\ 38.0\mathrm{m}=\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2 \\ \mathrm{t}=2.78\mathrm{s} \end{gathered}$$ 

 

So she has been jogging for  

                                                                                           

$9.00\mathrm{s}+2.78\mathrm{s}=11.78\mathrm{s}$

 

The horizontal distance with this time she travels is,

$$\begin{gathered} s=ut \\ s=\left(3.05m/s\right)\left(11.78s\right) \\ s=35.09m \end{gathered}$$                                                                                          

Bruce must throw the bagels so they travel horizontally about 35.9m in the time period of 2.78s. Now from the Newton’s law,

$$\begin{gathered} s=ut \\ 35.9m=u\left(2.78s\right) \\ u=12.9m/s \end{gathered}$$

Step 4: (b) The catching distance point

 Henrietta is at 35.9m distance from the building when she catches the bagels.