1. The height of the building
Tossed a ball up from the top of a building, it will travel 5.0 seconds before striking the ground. if it is thrown with 18m/s at an angle of 30°, find;
The height of the building
Its velocity just when it reaches the ground
The distance it falls far from the building
Break it down:
Initial velocity has vertical and horizontal components.
Thrown a ball up
The total time. Time interval for entire trip.
from the top of a building, it will travel 5.0 seconds
Initial velocity. V0 = 18m/s
before striking the ground. if it is thrown with 18m/s
ɵ = 30°. Initial velocity vector deviate 30° from +x axis. We select the origin of our coordinate system at the point that the ball is tossed upward.
at an angle of 30°,
find;
Y =? Look at the diagram.
a) The height of the building
Determine both magnitude and direction.
b) Its velocity just when it reaches the ground
X =? Look at the diagram.
c) The distance it falls far from the building
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2. The height of the building
Tossed a ball up from the top of a building, it will travel 5.0 seconds before striking the ground. if it is thrown with 18m/s at an angle of 30°, find;
The height of the building
Its velocity just when it reaches the ground
The distance it falls far from the building
Solution:
+ y
V0x
Draw a diagram
V0y
V0 = 18 m/s
θ = 30°
+ X
Y = ?
X =?
V =?
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3. The height of the building
Tossed a ball up from the top of a building, it will travel 5.0 seconds before striking the ground. if it is thrown with 18m/s at an angle of 30°, find;
The height of the building
Its velocity just when it reaches the ground
The distance it falls far from the building
Solution:
Calculate the horizontal and vertical components of the initial velocity +Y
Sin30° = 0.5
Cos30° = 0.86
V0y= V0 sin30
 V0y =18 * 0.5
 V0y = 9.0 m/s
V0 = 18 m/s
θ = 30° +X
V0x = V0 cos30
 V0x =18 * 0.86
 V0x = 15.6 m/s
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4. y = [V0y * t] - [g * t2] /2  y = - 77.5m
Tossed a ball up from the top of a building, it will travel 5.0 seconds before striking the ground. if it is thrown with 18m/s at an angle of 30°, find;
The height of the building
Its velocity just when it reaches the ground
The distance it falls far from the building
Solution:
The height of the building
y = [V0y * t] - [g * t2] /2
 y = [9.0 * 5] - [9.8 * 25] /2
 y = m
The height of the building is 77.5m and negative sign shows it is measured below the origin which is at tossing point.
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5. Magnitude of velocity vector when striking the ground.
Tossed a ball up from the top of a building, it will travel 5.0 seconds before striking the ground. if it is thrown with 18m/s at an angle of 30°, find;
The height of the building
Its velocity just when it reaches the ground
The distance it falls far from the building
Solution:
b) Its velocity just when it reaches the ground
Velocity in any point can be calculated by V2 = Vx2 +Vy2.
We know VX is constant during the entire trip and equal to V0X.
 Vx = V0x = 15.6m/s
And to calculate Vy in any point we use Vy = V0y – [g * t] where t is the time that the object reaches that point, in this case ground level.
Negative sign shows that the direction of the vertical component of the velocity vector when it strikes the ground points along –y axis.
Vy = V0y – [g * t]
 Vy = 9 – [9.8 * 5]
 Vy = - 40m/s
V2 = Vx2 +Vy2
Magnitude of velocity vector when striking the ground.
 V2 = [15.6]2 +[-40]2
 V = 42.93m/s
tg α = Vy / Vx = - 40 /15.6
Velocity vector is 68° deviated below +x axis when striking the ground. Draw the components of the velocity vector at the ground level to better understand the direction of the velocity vector.
 α ≈ - 68°
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6. The height of the building
Tossed a ball up from the top of a building, it will travel 5.0 seconds before striking the ground. if it is thrown with 18m/s at an angle of 30°, find;
The height of the building
Its velocity just when it reaches the ground
The distance it falls far from the building
Solution:
c) The distance it falls far from the building
Horizontal motion is a motion with constant speed of Vx = V0x = +15.6m/s.
X = VX * t
 X = * 5
The distance that ball falls from the building is 78m.
And the horizontal displacement is along +x axis;
X =78m and X = +78m
 X = +78m
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