Proj Motion EX 3

Helena Dedic

Amorosa, who is on a balcony 40 m above the ground, throws the key to her heart at 37° below the horizontal to Pedro on the ground. He catches the key two seconds after it was thrown.

a. How far from the base of the building was he standing?

b. At what angle was the key moving when he caught it?

Solution:

It takes 2 s for the keys to reach Pedro. We will use the equations governing projectile motion, substituting t = 2 s :

1. $\Delta x=v_{x0}t$

2. $\Delta y=v_{y0}t-4.9m/s^{2}t^{2}$

3. $v_{y}=v_{y0}-9.81m/s^{2}t$

4. $-19.6m/s^{2}\Delta y=v_{y}^{2}-v_{y0}^{2}$

a. to find Δ x, we must know $v_{0}$.

We find $v_{y0}$ from equation 2 with t = 2 s

$v_{y0}(2s)=\Delta y+4.9m/s^{2}(4s)^{2}$ 

$v_{y0}(2s)=-10.2m/s$ 

Knowing $v_{y0}=v_{0}sin(-37^{o})$, we can find $v_{0}$

$v_{0}=10.2m/s/0.6=17m/s$ 

Now we can use equation 1 to find \$Deltax$

$\Delta x=v_{0}cos(-37^{o})t$ 

$\Delta x=-10.2m/s-9.8m/s^{2}(2s)=29.8m/s$ 

b. To find the angle that the keys are travelling when caught, we must find the components of their velocity:

$v_{x}=v_{0}cos(-37^{o})=13.6m/s$ 

$v_{y}=v_{y0}-9.8m/s^{2}t$ 

$v_{y}=-10.2m/s-9.8m/s=-29.8m/s$ 

• Back to Projectile Motion
  
  
• Back to Exercises on Projectile Motion