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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.°
V_{i} = 100. m/s
V_{iy} = ?
V_{iy}
= V_{i}(sinӨ)
= (100. m/s)sin60.°
V_{iy} = 87. m/s up
(b) Find
the objects
maximum height.
V_{iy} = 87. m/s
a_{y} = 10. m/s^{2} (ascending)
d_{y} (max) =?
V_{fy} = 0 (at
max height)
V_{f}^{2} = V_{iy}^{2}
+ 2ad
0^{2} = (87. m/s)^{2}
+ 2(10. m/s^{2})d
0 = 7569 + 20.d_{y}
subtract 7569 from both sides
7569 = 20.(d_{y})
d_{y} = 378 m
Actually ......
d_{y} = 380 m
(2 sig. figs)
(b) Total time in the air
V_{iy} = 87. m/s up
t = 2(87 m/s)/9.8 m/s^{2}
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.°
V_{i} = 40. m/s
V_{iy} = ?
V_{iy} = VsinӨ
= (40. m/s)sin45.°
(b) If the
rock strikes ground in 1.0 sec what
is
height of cliff?
d_{y} = ?
(height)
a_{y} = + 10. m/s^{2}
t = 1.0 sec
V_{iy} = 28 m/s
(from before)
d_{y} = V_{iy}t + (˝)at^{2}
= 28.m/s(1sec)+˝(10. m/s^{2})(1.0sec)^{2}
28 m + 5.0 m
d_{y} = 33 m down
At Peak,
V_{x}
= ?
V_{x}
= 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.
d_{x} = V_{x}t = Vcosqt
Need to find t
d_{y} = V_{iy}t + (˝)at^{2}
V_{iy }= Vsinq
V_{iy }= 8.6 m/s
d_{y} = V_{iy}t + (˝)at^{2}
2.20 m = (8.6m/s)t + 4.9t^{2}
0 = 4.9t^{2} +(8.6m/s)t  2.20m
Quadratic Equation
t = 1.99 s, 0.225 s
d_{x} = V_{x}t = Vcosqt
d_{x} = V_{x}t
= (15m/s)cos34(1.99 sec)
= 25.6 m
The ammunition fired from an M16 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.
V_{x} = VcosӨ = 600 m/s
d_{x} = VcosӨt = [600m/s]t =
Need to find t
Make down positive
Δd_{y} = VsinӨt + 1/2gt^{2}
50 m = 1000m/s[sin53.1]t + 4.9t^{2}
50 m = 800t + 4.9t^{2}
0 = 4.9t^{2 } 800t  50. m
t = 163 sec
V_{x} = VcosӨ = 600 m/s
d_{x} = VcosӨt = [600m/s]t =
d_{x} = 98000 m
(B) With what speed would the bullet hit?
V_{x} = 600 m/s
down Neg.
V_{fy} = VsinӨ + gt
V_{fy} = 800m/s + 10m/s^{2}(163sec)
V_{fy} = 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?
D_{x} = V_{x}t
500 m = [(1000m/s)cosӨ]t
Δd_{y} = V_{iy}t + (˝)at^{2}
0 = V_{iy}t + (˝)at^{2}
0 = VsinӨt + 5t^{2}
(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
Intro to
Circular Motion
ŠTony Mangiacapre.,
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