A golf ball with a velocity 40 at an angle 45 degree speed of bthe ball after 2 sec

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 =                          [taking the LCM]

v = 8.28 m/s                    [√2 = 1.41]

i think it would be 5