Projectile Motion

Projectile Fired at an Angle - AP Problems

AP-level problems: deriving time of flight and solving full angled launches

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AP-Level Projectile Problems — derive time of flight for a level-ground trajectory, then solve a full angled-launch word problem

Projectile Motion

Show Me The Physics Website

Graph of a projectile's Vx and Vy components over time

Projectile

Where 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

Time of Flight — Level Ground

Given

A projectile is fired at an angle θ and a velocity V. The object returns to level ground.
Find: express the time it takes for the object to hit the ground in terms of V, θ, and any other relevant variables.

↓

Viy = Visinθ

Vfy = -Visinθ

Vfy = Viy + at

-Visinθ = Visinθ - gt

Total Time — Level Ground

t = (2Visinθ)/g

4. Word Problems

www.sihunt.co.uk — Russian long jumper Tayana Lebedeva performs a double hitch action in the 2007 world finals in Osaka, Japan.

Ex) Long Jump

Given

An athlete doing a running jump leaves the ground at an angle of 25° and a velocity of 10. m/s.
(a) How long does it take the athlete to reach her maximum height?

↓
Diagram of the running jump launch angle

Viy = Visinθ = (10. m/s)sin25.

Viy = 4.2 m/s up

XY
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

-4.2 m/s = (-9.8 m/s2)t   (subtract 4.2 m/s from both sides)

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

t = .42 sec to reach peak

Find

(b) How far did she jump?

↓

On level ground: Total Time = double the peak time

Total Time = .84 seconds

dx = Vxt   (must find Vx first)

Vx = Vcosθ = (10. m/s)cos25.

Vx = 9.1 m/s

dx = 9.1 m/s(.84 sec)

dx = 7.6 m

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