Projectile Motion

What Forces are acting on a Projectile?

• Initial Force that caused motion• Force of gravity

• Gravity causes the object to curve downward in a parabolic path (trajectory)

• An Object’s motion can be broken down into it’s horizontal and vertical component vectors.– x and y vectors

• Important Rule: Horizontal motion does NOT affect vertical

motion!!!!

Vx is constant and there is 0 acceleration!

Vy is changing and acceleration is due to gravity.

• At the top of a path, – there is no y – component– Vx component only!!!

Projectile Problem Solving

• Problems in which an object was dropped with a force in the x- axis

V0

dy

dx

Projectile Problem Solving

• Problems in which an object was dropped with a force in the x- axis

From Free Fall dy = ½ gt2

V0

dy

dx

Projectile Problem Solving

• Problems in which an object was dropped with a force in the x- axis

From Free Fall dy = ½ gt2

From Linear Motiondx= v0t and that v0= dx / t

V0

dy

dx

Ex. 1 An object travelling at 50 m/s falls out of a plane. It hits the ground 10 s later. What is the

horizontal distance travelled by the object?

Ex. 1 An object travelling at 50 m/s falls out of a plane. It hits the ground 10 s later. What is the

horizontal distance travelled by the object?

v0 = 50 m/s t = 10 s dx= ?

Solving for the x-axis vector component• dx= v0t

Ex. 1 An object travelling at 50 m/s falls out of a plane. It hits the ground 10 s later. What is the

horizontal distance travelled by the object?

Solving for the x-axis vector component• dx= v0t

• dx= (50 m/s)(10s)

• dx= 500 m

** Ch. 3 problem 41 in HW

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m

• dx= 10 m

• v0 is unknown

v0= dx / t

But wait, t is unknown…..

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m• dx= 10 m • v0 is unknown

v0= dx / t

But wait, t is unknown….. And we can solve for it using dy = ½ gt2

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m

• dx= 10 m

• v0 is unknown

v0= dx / t

dy = ½ gt2 OKAY – Solve for time

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m

• dx= 10 m

• v0 is unknown

v0= dx / t

dy = ½ gt2 OKAY – Solve for time

50 m = ½ (10 m/s2) (t2) 10 s2 = t2 take the square root of both sides

t = 3.16 s

Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the

object initially launch in order to reach the pool?

• dy = 50 m

• dx= 10 m

• v0 is unknown

v0= dx / t

Knowing that t = 3.16 s, we can now solve for V0.

v0= dx / t = 10 m / 3.16 s = 3.16 m/s** problems 42 and 44 in the hw

Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below

the target does the arrow hit?

Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below

the target does the arrow hit?

• The only known given is time and we are determining the distance in the vertical direction dy.

• Which equation should we use????

Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below

the target does the arrow hit?

• The only known given is time and we are determining the distance in the vertical direction dy.

• Which equation should we use????• dy = ½ gt2

Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below

the target does the arrow hit?

• The only known given is time and we are determining the distance in the vertical direction dy.

• Which equation should we use????• dy = ½ gt2

• dy = ½ (10 m/s2) (0.2 s)2

• dy = 0.2 m ** problem 43

THE END!!!!