Solved problems in projectile motion – determine the position of an object

1. A body is projected upward at an angle of 60^{o }to the horizontal with an initial speed of 12 m/s. Determine the position of the object after moving for 1 second! Acceleration of gravity is 10 m/s^{2}.

__Known :__

Angle (θ) = 60^{o}

Initial velocity (v_{o}) = 12 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s^{2}

__Wanted :__ Object position after moving during 1 second

__Solution :__

Horizontal component of initial velocity :

v_{ox }= v_{o} cos θ = (12 m/s)(cos 60^{o}) = (12 m/s)(0.5) = 6 m/s

Vertical component of initial velocity :

v_{oy }= v_{o} sin θ = (12 m/s)(sin 60^{o}) = (12 m/s)(0.5√3) = 6√3 m/s

**Object position at horizontal direction:**

__Known :__

Horizontal component of velocity (v_{x}) = 6 m/s

Time interval (t) = 1 second

__Wanted :__ horizontal range (x)

__Solution :__

6 meters / second means the ball moves as far as 6 meters every 1 second. The distance of the ball after moving for 1 second is 6 meters. So the position of the ball in the horizontal direction is 6 meters.

**Object position at vertical direction :**

Choose upward direction as positive and downward direction as negative.

__Known :__

Initial velocity (v_{o}) = 6√3 m/s (positive upward)

Time interval (t) = 1 second

Acceleration of gravity (g) = -10 m/s^{2 }(negative downward)

__Wanted :__ height after moving during 1 second

__Solution :__

h = v_{o }t + 1/2 g t^{2} = (6√3)(1) + 1/2 (-10)(1^{2}) = 6√3 + (-5)(1) = 6√3 – 5 = 6(1.7) – 5 = 10.2 – 5 = 5.2 meters.

**Object position after moving for 1 second :**

Horizontal displacement (x) = 6 meters

Vertical displacement (y) = 5.2 meters

[irp]

2. A body is projected upward at an angle of 30^{o }to the horizontal from a building 20 meters high. Its initial speed is 50 m/s. Calculate the vertical displacement after the body moving for 1 second! Acceleration of gravity is 10 m/s^{2}.

__Known :__

Angle (θ) = 30^{o}

Initial height (h_{o}) = 20 meters

Initial velocity (v_{o}) = 50 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s^{2}

__Wanted :__ Height (h)

__Solution :__

**Vertical component of initial velocity :**

v_{oy} = v_{o} sin θ = (50 m/s)(sin 30^{o}) = (50 m/s)(0.5) = 25 m/s

**Height :**

Choose upward direction as positive and downward direction as negative.

__Known :__

Initial velocity (v_{o}) = 25 m/s (positive upward)

Time interval (t) = 1 second

Acceleration of gravity (g) = -10 m/s^{2} (negative downward)

__Wanted :__ Height (h)

__Solution :__

h = v_{o }t + 1/2 g t^{2 }= (25)(1) + 1/2 (-10)(1^{2}) = 25 + (-5)(1) = 25 – 5 = 20 meters.

The height of the body after moving for 1 second is 20 meters above where the body is projected or 40 meters above ground.

[irp]

3. A small ball projected horizontally with initial velocity v_{o} = 10 m/s from a building 10 meters high. Calculate the ball displacement after moving 1 second! Acceleration of gravity is 10 m/s^{2}

__Known :__

Initial height (h) = 10 meters

Initial velocity (v_{o}) = 10 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s^{2}

__Wanted:__ Position of the ball after moving 1 second!

__Solution :__

**Horizontal displacement :**

__Known :__

Horizontal component of velocity (v_{x}) = 10 m/s

Time interval (t) = 1 second

__Wanted:__ Position of the object

__Solution :__

10 meters/second means the object move as far as 10 meters every 1 second. Displacement after moving for 1 second is 10 meters. So horizontal displacement is 10 meters.

**Vertical displacement :**

*Calculated as the free fall motion.*

__Known :__

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s^{2 }

__Wanted :__ Height after moving during 1 second (h)

__Solution :__

h = 1/2 g t^{2} = 1/2 (10)(1^{2}) = (5)(1) = 5 meters.

After 1 second, the object falls as far as 5 meters. Height above the ground level = 10 meters – 5 meters = 5 meters.

**The position of the object after moving 1 second :**

Position of the object at horizontal direction (x) = 10 meters

The position of the object at vertical direction (y) = 5 meters

[irp]

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- Resolve initial velocity into horizontal and vertical components
- Determine the horizontal displacement
- Determine the maximum height
- Determine the time interval
- Determine the position of the object
- Determine the final velocity