The football has 2 velocities.
Which velocity is the football player matching.

 

 

 

 

Horizontal velocity.
The football and the football player
have a constant horizontal velocity.

 

 

St. Mary's Show Me The Physics

 

 

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

(a) Find the objects maximum height & total time in the air

(level ground)

 

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

 

First find the initial vertical component of the object's velocity 

 

Ө = 60.°

 

Vi  = 100. m/s
Viy = ?

 

 

Viy = Vi(sinӨ)

 

 

= (100. m/s)sin60.°

 

 

 

 

Viy = 87. m/s up

 

 

 

 

 

(b) Find the objects maximum height.

 

Viy = 87. m/s
ay = -10. m/s2 (ascending)
dy (max) =?
Vfy = 0 (at max height)

 

 

Vf2 = Viy2 + 2ad

 

 

02 = (87. m/s)2 + 2(-10. m/s2)d

 

0 = 7569 + -20.dy

 

 

subtract 7569 from both sides

 

 

-7569 = -20.(dy)

 

 

dy = 378 m

 

 

Actually ......

 

 

 

dy = 380 m

 

 

(2 sig. figs)

 

(b) Total time in the air

Viy = 87. m/s up

 

t = 2(87 m/s)/9.8 m/s2

 

t = 18 sec


 

 

 

 

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

(a) If rock strikes ground in 1 sec, height of cliff?

        

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

Ө = 45.°

 
Vi = 40. m/s
Viy = ?     
 
Viy = VsinӨ
 
= (40. m/s)sin45.° 

			
 
Viy = 28 m/s down 

 

 

 

 

 

 

 

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

 

 

dy = ? (height)

ay = + 10. m/s2

 

t = 1.0 sec

 

Viy = 28 m/s (from before)

 

 

 

dy = Viyt + (˝)at2
 
 
 
 
 
= 28.m/s(1sec)+˝(10. m/s2)(1.0sec)2
 
 
 
28 m + 5.0 m
 

		
 
dy = 33 m down
 
 
 
 

 

At Peak,

 

 Vx = ?

 

 

 

Vx = 9.0 m/s

 

 

 

 

h

Total Time in Air?

 

 

 

1.84 sec
(twice peak)

 

 

actually 1.8

 

 

 

(2 sig. figs.)

 

 

 

 

 

What will the horizontal velocity of the baseball be just before it hits the ground?

 

 

 

 

9 m/s, Vx is constant

 

Ex) A wild bowler releases the his bowling ball at a speed of 15 m/s and an angle of 34.0 degrees above the horizontal.

 

Calculate the horizontal distance that the bowling ball travels if it leaves the bowlers hand at a height of 2.20 m above the ground.

 

 

dx = Vxt = Vcosqt

 

Need to find t

 

dy = Viyt + (˝)at2

		
Viy = Vsinq
Viy = 8.6 m/s
 
dy = Viyt + (˝)at2
 
2.20 m = (-8.6m/s)t + 4.9t2
0 = 4.9t2 +(-8.6m/s)t - 2.20m
 
Quadratic Equation

				
t = 1.99 s, -0.225 s

 

dx = Vxt = Vcosqt

dx = Vxt

 

= (15m/s)cos34(1.99 sec)

 

=  25.6 m

 

The ammunition fired from an M-16 rifle has a muzzle velocity of 1000 m/s. A sniper perched on a tower 50 m high aims at a target and fires.

(A) The sniper aims her rifle 53.1 degrees above the horizontal. How far away from the platform does the bullet hit the ground?

Neglect air resistance.

 

Vx = VcosӨ = 600 m/s

dx = VcosӨt = [600m/s]t =

 

Need to find t

 

Make down positive

Δdy = -VsinӨt + 1/2gt2

50 m = -1000m/s[sin53.1]t + 4.9t2

50 m = -800t + 4.9t2

 0 = 4.9t2 - 800t - 50. m

 

t = 163 sec

 

Vx = VcosӨ = 600 m/s

dx = VcosӨt = [600m/s]t =

 

dx = 98000 m

 

(B) With what speed would the bullet hit?

Vx = 600 m/s

down Neg.

Vfy = -VsinӨ + gt

Vfy = -800m/s + 10m/s2(163sec)

Vfy = 830 m/s

 

Pythagorean Theorem

 

V = 1000 m/s

 

(C) If the target is 500 meters away on a platform 50 meters high, what angle would the sniper have to make with her gun to hit the target?

 

Dx = Vxt

500 m = [(1000m/s)cosӨ]t

 

Δdy = Viyt + (˝)at2

 

0 = Viyt + (˝)at2

0 = VsinӨt + 5t2

 

 

 

 

 

 

(A) 98000 m (B) 1000 m/s (C) 44.9

 


Monkey and Hunter
 Don Ion

(offsite)

 

 

 Flash Physics
 Don Ion

 

 

St. Mary's U. Astronomy and Physics Dept.

| Onsite |

 

 

 

 

 

 

 

 

 

 

Projectile Motion Game
St. Mary's Physics

 

 

 

 


 

 

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