Nuzhat at 19:52, 16 August 2010

2010-08-16T19:52:04Z

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''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:''' 

[[image: Proj Motion Sol 3.png|TOP]] 

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

1. <math>\Delta x = v_{x0}t</math> 

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

3. <math>v_y = v_{y0} - 9.81m/s^2 t</math> 

4. <math>-19.6m/s^2 \Delta y = v_y^2 - v_{y0}^2</math> 

a. to find Δ x, we must know <math>v_0</math>. 

We find <math>v_{y0}</math> from equation 2 with t = 2 s

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

<math>v_{y0}(2 s) = -10.2 m/s</math> 

Knowing <math>v_{y0} = v_0 sin (-37^o)</math>, we can find <math>v_0</math> 

<math>v_0 = 10.2 m/s / 0.6 = 17 m/s</math> 

Now we can use equation 1 to find \<math>Deltax</math> 
<math>\Delta x = v_0 cos(-37^o) t</math> 

<math>\Delta x = -10.2 m/s - 9.8 m/s^2 (2 s) = 29.8 m/s</math> 

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

<math>v_x= v_0 cos(-37^o)= 13.6 m/s</math> 

<math>v_y= v_{y0} - 9.8m/s^2t</math> 

<math>v_y= -10.2 m/s - 9.8 m/s = -29.8 m/s</math> 

[[image: Proj Motion Sol 3b.png|TOP]] 

*[[Projectile Motion|Back to Projectile Motion]] 
*[[Exercises on Projectile Motion|Back to Exercises on Projectile Motion]]

Proj Motion EX 3 - Revision history

Nuzhat at 19:52, 16 August 2010