PROJECTILE MOTION

PROJECTILE MOTION

Unit 4 Module 1 Science 9

PRAYER

Poll

ATTENDANCE

Multiple Choice

Which component of projectile is constant?

• horizontal velocity

• vertical velocity

• range

• angle of trajectory

Multiple Choice

What is the curved path taken by the projectile motion?

• trajectory

• horizontal path

• upward path

• downward path

Multiple Choice

which described as the combination of a constant horizontal motion and free fall?

• inertia

• projectile

• momentum

• energy

Multiple Choice

What is the horizontal distance traveled by a projectile motion?

• height

• trajectory

• projectile

• range

Multiple Choice

At what angle should a water hose be intended for water to land with the greater range?

• 0⁰

• 30⁰

• 45⁰

• 60⁰

Multiple Select

Put check in the sports/situation illustrating a curved motion.

• Throwing a basketball into the basket

• Kicking a soccer ball

• Batting a baseball

• Rolling a ball in the ground

• Dropping a ball from the building

PROJECTILE MOTION

The combination of vertical and horizontal motion is acted by gravity only.

TRAJECTORY

The curved path is taken by the object.

HEIGHT

The vertical distance the ball can reach.

RANGE

The horizontal distance along the ground.

HORIZONTAL VELOCITY

The component of velocity in projectile motion that is constant.

VERTICAL VELOCITY

The velocity decreases when moving upward and increases when moving downward.

Open Ended

Problem 1. A ball is thrown horizontally with a speed of 2.50 m/s from a height of 50.0 meters. Find (a.) the range of the projectile. (b.) how long will the ball traveled before it reaches the ground?

Answer Problem 1

Given: Vx = 2.50m/s; h= 50.0 m; g= 9.8 m/s2

Find: Range and time

Formula: R = Vxt          ;          t = $\sqrt{\frac{2h}{g}}$  
A. Range (R) = 7.975 or 8 meters
B. time (t)= 3.19 seconds

Open Ended

Problem 2. Am object is thrown horizontally from the top of a building with an initial velocity of 15.0 m/s. Find the height of the building if the object hit the ground 60.0 meters away from the building.

Answer Problem 2

Given: Vy = 15.0 m/s ; R = 60.0 m ; g= 9.8 m/s²

Find: height

Formula: h = gt²  / 2 ; Vy = gt or t = Vy / g

Answer : t = 1.53 seconds ; h = 11.47 meter

PROJECTILE AT AN ANGLE

formula

Vx = Vxi = cos  $\theta$  
Vy = Vi sin  $\theta$  

Open Ended

Problem 3. What must be the horizontal distance of the ball if it is thrown with initial velocity of 50.0 m/s at an angle of 20⁰?

Given Vi=50.0 m/s     θ = 200

Unknown H =?

Formula: H = gt² / 2 ; V_y = V_i sinθ

 

h= (V_i sin ) (t) - (1/2) gt²

R = V_i2 sin2θ / g

Solution R = (50m/s)² sin2 (20) / 9.8 m/s²

Answer: R = 163.97 m or 164 m

WHAT I CAN DO

WHAT I CAN DO: STEPS

• 1. Vy= gt ( Vertical Velocity)

• 2. H = gt² (Vertical Displacement)

• 3. V_x = 30 m/s ( given on the problem horizontal speed 30m/s) - (Horizontal Velocity)

• 4. R = V_xt ( Horizontal Displacement)