Projectile Motion Vy,Vx  - St. Mary's H.S. Physics

Projectile Motion - St. Mary's H.S. Physics

 

Vx (m/s) Vy
(m/s)
25 40
25 30
25 20
25 10
25 0 (peak)
25 -10
25 -20
25 -30
25 -40

 

 

4. Word Problems

Ex) An athlete doing a running jump leaves ground at an angle of 25.° and a velocity of 10. m/s.

(a) What is the initial vertical component (Viy) of the athlete's velocity?

 

 

Ex)  An athlete doing a running jump leaves ground at an angle of 25.° and a velocity of 10. m/s.

 

 

(a) What is the initial vertical component (Viy) of the athlete's velocity?

                   

 

 

 

Viy = Vsinq

 

 = (10. m/s)sin25.

 

 

 

= 4.2 m/s

 

 

 

(b) How long does it take for athlete to reach her maximum height?

 

 

 

 

 

 

 (b) How long does it take for athlete to reach her maximum height?

 

 

X Y
t = ? Viy = 4.2 m/s
(from above)
 
t  =  ?
  Vfy= 0
(peak)
 
ay = -9.8 m/s2

 

 

Vf = Vi + at

0 = 4.2 m/s + (-9.8 m/s2)t

 

- 4.2 m/s = (-9.8 m/s2)t

 

(- 4.2 m/s)/(-9.8 m/s2) = t

 

t = .43 secs
 to reach peak

 

 

 

(b) How long did it take for the athlete to complete the entire jump?

 

 

 

 

(b) How long did it take for the athlete to complete the entire jump?

 

 



Total time = double peak time

 

 

 

Entire Jump:

total time = .86 seconds

 

 

 

(c) How far did she jump?

 

dx = ?

 

(c) How far did she jump?

 

dx = ?

t = .86 sec

 

dx = Vxt

 

 

(Must find Vx first)

 

Vx = Vcosq

 

Vx = (10. m/s)cos25.

 

Vx = 9.1 m/s

 

Vx = Vcosq

 

dx = 9.1 m/s(.86 sec)

 

 

 

dx = 7.8 m

 

 

 

Ex) A projectile leaves ground at an angle of 60° and a speed of 100m/s.

 

(a) Find the initial vertical component of the object's velocity 

 

 

Ex) A projectile leaves ground at an angle of 60.° and a speed of 100. m/s.

 

(a) Find the initial vertical component of the object's velocity 

 

 

q = 60.°

 

Vi  = 100. m/s

 

 

Viy = Vi(sinq)

 

 

= (100. m/s)sin60.° (vertical component)

 

 

 

 

Viy = 87. m/s

 

 

 

(b) Find the objects maximum height.

 

Viy = 87. m/s
ay = -9.8 m/s2
dy (max) =?
Vfy = 0 (at max height)

 

 

Vf2 = Viy2 + 2ad

 

 

02 = (87. m/s)2 + 2(-9.8 m/s2)d

 

0 = 7569 + -19.6d

 

 

-7569 = -19.6d

 

 

dy = 383m

 

or
 378 m (If a =10 m/s2)

 

 

Ex) A rock is thrown from a cliff with initial speed of 40. m/s at an angle of 45.° below  the horizontal. 

a) What is the vertical component of the initial velocity?

 
q = 45°
 
Vi = 40 m/s
     
 
Viy = Vsinq
 
= (40. m/s)sin45.° 

			
 

			
Viy = 28 m/s

		

		
 

		
(b) If the rock strikes ground in 1 sec what is height of cliff?
		

		
 

		
 

		
 

		
(b) If the rock strikes ground in 
		1.0 sec what is 
		height of cliff?

		
 

		
a = + 9.8 m/s2

		
t = 1.0 sec

		
Viy = 28 m/s (from before)

		
dy = ?

		
 

		
dy = Viyt + (½)at2

		
 

		

			

				

		
= 28 m/s(1sec)+½(9.8m/s2)(1 sec)2

				

			

		

		
 

		
28 m + 4.9 m

		
 

		
dy = 32 m

		

		
 

		

		

		

		 

		

		 

		

		
At peak, Vy = ?, Vx 
		= ?

		
 

		
 

		
 

		
 

		
Vy = 0, Vx 
		= 9 m/s

		
 

		

		
 

		
 

		
h

		

		 

		

		total time in air?

		

		 

		

		 

		

		 

		

		1.84 sec (twice peak)

		

		 

		

		

		 

		

		

		

		dx = ? 

		

		 

		

		dx = Vxt

		

		 

		

		

		
dx = (9.0 m/s)1.84 
		sec

		

		
dx = 16.56 m or 17 m

		

		

		
 

		

		Monkey 
		and Hunter

 - Don Ion -
		Flash Physics

		

		 

		

		 

		

		 

		

		

		

		 

		

		 

		

		

		

		 

		

		

		 

		

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