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You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time you drop your ball from the window. The two balls are initially separated by 28.7 m. (a) At what time do they pass each other? (b) At what location do they pass each other relative the window?

Respuesta :

syed514
Let they will meet at y above surface ground. 
Vf² = v ² + 2 g H 
Vf² = 0 ² + 2 (9.8)(28.7) 
Vf = 23.7175 m/s 

analyzing for the first ball, 
y = Yo + ½ g t² 
y = 28.7 + ½ (-9.8) t² 


and for the second one, 
y = Vf t + ½ g t² 
y = 23.7175 t + ½ (-9.8) t² 


and then. they meet each other, 
y = y 
28.7 + ½ (-9.8) t² = 23.7175 t + ½ (-9.8) t² 
t = 1.21008 sec 

y = 23.7175 t + ½ (-9.8) t² 
y = 23.7175 (1.21008) + ½ (-9.8)(1.21008)² 
y = 21.5250 m
shubhamchouhanvrVT

(a) The time at which the two balls will meet [tex]t=1.21\ sec[/tex]

(b) The position at which two balls will meet [tex]y=21.52\ m[/tex]

What is equation of motion?

The equation of motions are the equations which defines the characteristics of any object which are in motion like the velocity, position, time etc

Now it is given in the question that

Height from which the ball dropped H=28.7 m

Let they will meet at y above surface ground. 

[tex]V_f^2=v^2+2gH[/tex]

[tex]V_f^2=0+2\times 9.81\times (28.7)[/tex]

[tex]V_f=23.71\ \frac{m}{s}[/tex]

Now the position of the first ball

[tex]y=y_o+\frac{1}{2}(gt^2)[/tex]

[tex]y=28.7+\frac{1}{2}\times (-9.81)\times t^2[/tex]

Now for the second ball the position is

[tex]y=V_ft+\frac{1}{2}gt^2[/tex]

[tex]y=23.71+\frac{1}{2}(-9.81)t^2[/tex]

On the point of meet of the two balls the the displacement will be same

y = y 

[tex]28.7+\frac{1}{2}(-9.8)t^2=23.71t+\frac{1}{2}(-9.8)t^2[/tex]

[tex]t=1.21\ sec[/tex]

Now the position at which two balls meet will be

[tex]y=23.7175t+\frac{1}{2}(-9.8)t^2[/tex]  

[tex]y=23.71(1.21)+\frac{1}{2}(-9.81)(1.21)^2[/tex]

[tex]y=21.52\ m[/tex]

so the time at which the two balls will meet [tex]t=1.21\ sec[/tex] and the position at which two balls will meet [tex]y=21.52\ m[/tex]

To know more about equation of motion follow

https://brainly.com/question/25951773

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