Projectile Motion
Derivation of the Projectile
Equations
Vx=cosӨ
V0
Ө
Vy=V0sin Ө - gt
V0
Ө
h=v02sin2Ө
2g
V2 = V02 + 2ax
0 = V02 + 2ax
0 = V02 + 2ah
-V02 = h
2a
V02 = h
2g
h=v02sin2Ө
2g
T=2V0sinӨ
g
v = v0+at
t = v/a
t = v/g
t = 2v/g
T = 2V0sinӨ
g
R=V02sin2Ө
g
X=VT
T=2V0sinӨ
V=V0cos Ө
X=2v0sin Ө V0cos Ө/g
R=V02sin2Ө
g
x = (V0cosӨ)t
x = vt
V=V0cosӨ
X = (V0cosӨ)t
Y=(V0sinӨ)t – ½ gt2
Y = vt + ½ at2
Y = (V0sinӨ)t – ½ gt2
The big trajectory Equation
y = (tanӨ)x – gX2/ 2V02cos2Ө
X = vt + ½ at2
Y = vt + ½ at2
T=X/V0cosӨ
X = V(X/V0cosӨ)+ ½ a[X/V0cosӨ]2
X = V(X/V0cosӨ)- ½ g[X/V0cosӨ]2
V=V0sinӨ
X = V0sinӨ(X/V0cosӨ)- ½ g[X/V0cosӨ]2
X = (tan Ө)x – gx2
2(V02cos2Ө)
Find the initial vertical and
horizontal velocity.
The initial velocity is 30 m/s. The angle is
40º. What is the initial horizontal and
vertical velocity?
V0
Ө
Answers
Use the Vx=cosӨ and Vy=V0sin Ө - gt to get
Vx = 23.0 m/s
Vy= 19.3 m/s
Height, Time and Range
The initial velocity is 30 m/s. The angle is
40º. What is the height the projectile
attains, how long is it in the air and how far
does it go?
Answers
Use h=v02sin2Ө
2g to get 19.0 m for the height.
Use T=2V0sinӨ
g
to get 3.9 sec for the time
Use R=V02sin2Ө
g to get 90.4 m for the range.
Jumping Angles
Greta Greyhound is jumping hurdles. She is
running a velocity of 18.3 m/s and the
hurdles are .75 m high. At what angle
must she leave the ground in order to
clear the hurdles?
Answer
Use the height equation and solve for θ.
h = Vo2 sin2 θ
2g
w 2gh = sinθ
Vo
sin-1ow2gh p = θ
o Vo p
θ = 12.1 degrees
Go, Greta!!!
References
• Schaum’s Outlines Applied Physics Fourth
Edition Arthur Beiser, Ph.D. McGraw-Hill
2004 p 42.