1. Angled Projectiles

4. Consider a projectile with initial velocity vi at angle :
Let’s find the x and y components of vi vi
viy 
vix
vix = vicos
(determines the x-motion)
viy = visin
(determines the y-motion)
Acceleration in the
y-direction due to gravity
(ay = -9.8 m/s2)
Acceleration in the
x-direction = 0
(ax = 0)
Equations:
vy = viy + ayt
y = viyt + ½ ayt2
vy2 = viy2 + 2ayy
y = ½ (viy+vfy)t
Velocity in the
x-direction is constant
vx = vix the whole time

7. Ex:
If a projectile is launched from ground level with an initial velocity of 20 m/s at an angle of 60 from the horizontal:
Find vix and viy.
How high does it go?
Find the total time in the air.
How far from the launch point does it land?

8. PhET projectile Motion applet

9. Projectiles range angle applet

10. *
*Use only for the following situation: v0  R
Which angle  gives the maximum range?
How does the range compare for complementary angles?

11. vox = 100 cos 30 = 86.6 m/s voy = 100 sin 30 = 50 m/s x = vox t
4. An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find:
a. Its position and velocity after 8 s
vox = 100 cos 30
= 86.6 m/s
voy = 100 sin 30
= 50 m/s
vo = 100 m/s, 30
t = 8 s
g = m/s2
x = vox t
= 86.6(8)
= m
y = voy t + ½ gt2
= 50(8) + ½ (-9.8)(8)2
= 86.4 m
vx = vox = 86.6 m/s
vy = voy + gt
= 50 + (-9.8)(8)
= m/s

12. At the top of the path: vy = 0 vy = voy + gt = 5.1 s Total time T = 2t
b. The time required to reach its maximum height
At the top of the path:
vy = 0
vy = voy + gt
= 5.1 s
c. The horizontal distance (range)
Total time
T = 2t
= 2(5.1)
= 10.2 s
x = vox t
= 86.6(10.2)
= m

13. time to maximum height = 1 s at the top vy = 0 vy = voy + gt
5. A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 s.
a. At what angle with respect to the horizontal was it released?
vo = 15 m/s
t = 2 s
time to maximum height = 1 s
at the top vy = 0
vy = voy + gt
= 40.8º

14. y = voy t +½gt2 = (15)(sin 40.8º)(1) + ½ (-9.8)(1)2 = 4.9 m
b. What was the maximum height achieved by the ball?
y = voy t +½gt2
= (15)(sin 40.8º)(1) + ½ (-9.8)(1)2
= 4.9 m

15. θ = 55º total time of flight: y = voy t +½gt2 if y = 0
6. An arrow was shot at an angle of 55º with respect to the horizontal. The arrow landed at a horizontal distance of 875 m. Find the velocity of the arrow at the top of its path.
θ = 55º
x = 875 m
total time of flight:
y = voy t +½gt2
if y = 0
0 = t (voy + ½ gt)
voy = ½ gt
vo sin θ =½ g t

16. substituting the time and solving for vo
= 95.5 m/s
At the top of its path the arrow has vy = o and
vx = vox
= 95.5 cos 55º
= 54.8 m/s

17. At top vy = 0 vy = voy + gt x = vxt = vo sin θ - gt Total time = 2t
7. Find the range of a gun which fires a shell with muzzle velocity vo at an angle θ . What is the maximum range possible?
At top vy = 0
vy = voy + gt
= vo sin θ - gt
x = vxt
Total time = 2t

18. sin θ cos θ= ½ sin 2θ
Maximum range is 45 since 2θ = 90