Answer:
Vertical velocity = 8.28 m/s
Horizontal velocity = 28.3 m/s
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We are Given:
Initial velocity of the ball = 40 m/s
Angle of elevation of the ball = 45°
Time interval = 2 seconds
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Vertical and Horizontal Components:
Vertical Component:
The vertical component of the ball can also be written as:
40*Sin(45°)
40*(1/√2) [since Sin(45°) = 1/√2]
40/√2 m/s
Horizontal Component:
The horizontal component of the ball can also be written as:
40*Cos(45°)
40*(1/√2) [Since Cos(45°) = 1/√2]
40/√2 m/s
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Speed of the Ball after 2 seconds:
Horizontal Velocity:
Since the horizontal velocity is not affected by any force during the flight, the horizontal velocity after 2 seconds will also be 40/√2 m/s
Since √2 = 1.41:
Horizontal velocity = 28.3 m/s
Vertical Velocity:
v = u + at [First equation of motion]
v = 40/√2 + (-10)(2)
[g is negative because it is acting against the direction of motion]
v = 40/√2 - 20
v = [tex]\frac{40-20\sqrt{2} }{\sqrt{2} }[/tex] [taking the LCM]
v = 8.28 m/s [√2 = 1.41]
