Slide 1

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```A soccer player kicks a ball into the air at an angle of 34.0°
above the horizontal. The initial velocity of the ball is +30.0
m/s. How long is the soccer ball in the air?
 = 34.0°
Vi = +30.0 m/s
Vy = Vi*sin
t=?
t = 2Vy/g
Vy = 30.0Sin(34.0) = 16.8 m/s
t = -2*16.8/-9.8 = 3.42 s
What is the horizontal distance traveled by a soccer ball
kicked into the air at an angle of 30.0° above the horizontal
with initial velocity of +31.0 m/s?
 = 30.0°
Vi = +31.0 m/s
Dx = ?
Vy = Vi*Sin
Vx = Vi*Cos
t = 2Vy/g
Dx = t*Vx
Vy = 31Sin(30) = 15.5
Vx = 31Cos(30) = 26.8
t = (-2*15.5)/-9.8 = 3.16
Dx = 3.16*15.5 = 84.6 m
What is the maximum height of a soccer ball kicked into the
air at an angle of 32.0° above the horizontal with the initial
velocity of +25.0 m/s?
= 32.0°
Vi = 25.0 m/s
Dy = ?
Vy = 25Sin(32) = 13.2 m/s
Dy = 13.22/(2*9.8) = 8.95 m
Vy = Vi*Sin
Dy = Vy2/2g
A ball falls from rest from a height of 452 m.
a. How long does it remain in the air?
b. If the ball has a horizontal velocity of 1.00*102 m/s when it
begins to fall, what horizontal displacement will it have?
Dy = 452 m
Vx =
1.00*102
t=?
Dx = ?
t2 = (2*452)/9.8 = √92.2 = 9.60 s
Dx = 9.60*100 = 9.60*102 m
t2 = 2Dy/g
Dx = t*Vx
An archer stands 40.0 m from the target. If the arrow is shot
horizontally with a velocity of 87.0 m/s, how far above the
bull’s-eye must she aim to compensate for gravity pulling her
arrow downward?
Dx = 40.0 m
Vx = 87.0 m/s
Dy = ?
t = 40.0/87.0 = .459 s
Dy = .5*9.8*.4592 = 1.04 m
t = Dx/Vx
Dy = .5gt2
A bridge is 142.9 m above a river. If a lead-weighted fishing
line is thrown from the bridge with a horizontal velocity of
24.0 m/s, how far has it moved horizontally when it hits the
water?
Dy = 142.9 m
Vx = 24.0 m/s
Dx = ?
t2 = (2*142.9)/9.8 = √29.2 = 5.40 s
Dx = 5.4*24.0 = 1.30*102 m
t2 = 2Dy/g
Dx = t*Vx
A beach ball, moving with a speed of +1.12 m/s, rolls off a
pier and hits the water 0.65 m from the end of the pier. How
high above the water is the pier?
Vx = 1.12 m/s
t = Dx/Vx
Dx = 0.65 m
Dy = .5gt2
Dy = ?
t = 0.65/1.12 = .58 s
Dy = .5*9.8*.582 = 1.7 m
A shot put is released with a velocity of 12 m/s and stays in
the air for 1.9 s.
a. At what angle with the horizontal was it released?
b. What horizontal distance did it travel?
Vi = 12 m/s
Vy = Vf-at
t = 1.9 s
θ = Sin-1(Vy/Vi)
θ=?
Vx = Vi*Cosθ
Dx = ?
Dx = Vx*t
Vy = 0-9.8*0.95 = 9.3 m/s
= Sin-1(9.3/12) = 51°
Vx = 12Cos(51) = 7.6 m/s
Dx = 7.6*1.9 = 14 m
A football is kicked at 45° and travels 82 m before hitting the
ground.
a. What is the initial velocity?
b. How long was it in the air?
c. How high did it go?
= 45°
Dx = 82 m
Vi = ?
t=?
Vi = √(g*Dx)/(2Sin*Cos)
Vx = Vi*Cos
t = Dx/Vx
Vy = Vi*Sin
Dy = Vy2/2g
Dy = ?
Vi = √(9.8*82)/[2Sin(45)Cos(45)] = 28 m/s
Vx = 28Cos(45) = 20 m/s
t = 82/20 = 4.1 s
Vy = 28Sin(45) = 20 m/s
Dy = 202/(2*9.8) = 20 m
A golf ball is hit with a velocity of 46.0 m/s at 32.0° above the
horizontal. Find
a. The range of the ball
b. The maximum height of the ball
Vi = 46.0 m/s
Range = Vi2Sin(2)/g
 = 32.0°
Dy = Vy2/2g
Range = 462Sin(2*32)/9.8 = 194 m
24.42/(2*9.8) = 30.3 m
```