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Projectile Fired at an Angle — word problems combining horizontal and vertical motion

Goalie catching a soccer ball

Which velocity, Vx or Vy, is constant
for a projectile?

Diagram of projectile Vx and Vy components

Projectile Motion — Volleyball

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Projectile Motion

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Vx and Vy components of a projectile over time

Sample Projectile

Vix = 25 m/s   |   Viy = 40 m/s

t (sec) Vx (m/s) Vy (m/s)
02540
12530
22520
32510
4250 (peak)
525-10
625-20
725-30
825-40

Word Problems

Example — Running Long Jump

An athlete doing a running jump leaves the ground at an angle of 25.° and a velocity of 10. m/s.

Part (a)

Given

Angle = 25.°, Velocity = 10. m/s
Find: (a) the initial vertical component (Viy) of the athlete's velocity.

Vector diagram of velocity components at an angle

Viy = Visinθ

= (10. m/s)sin25.°

Viy = 4.2 m/s up

Part (b)

Given

Viy = 4.2 m/s (from part a)
Find: (b) how long does it take for the athlete to reach her maximum height?

X Y
t = ? Viy = 4.2 m/s
(from before)
  t = ?
  Vfy = 0 (peak)
  ay = -9.8 m/s2

Vf = Vi + at

0 = 4.2 m/s + (-9.8 m/s2)t

(subtract 4.2 m/s from both sides)

- 4.2 m/s = (-9.8 m/s2)t

(- 4.2 m/s)/(-9.8 m/s2) = t

t = .43 sec to reach peak

Part (c) — Total Time

Given

Peak time = .43 sec, on level ground
Find: (c) how long did it take for the athlete to complete the entire jump?

On level ground: Total time = double peak time

Entire Jump:

Total Time = .86 seconds

Motion shot of the athlete's jump trajectory

Part (c) — Distance

Given

dx = ?, t = .86 sec
Find: (c) how far did she jump?

dx = Vxt  (must find Vx first)

Vx = Vcosθ

Vx = (10. m/s)cos25.°

Vx = 9.1 m/s

dx = 9.1 m/s(.86 sec)

dx = 7.8 m

Goalie catching a soccer ball