A marble is shot at an angle of 30 degrees and has an initial velocity of 40 m/s.  How many seconds to reach the ground and how far away does it land?

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The marble is shot at an angle of 30 degrees to the horizontal and has an initial velocity of 40 m/s. The initial velocity of the marble can be divided into 2 components. The horizontal component is equal to 40*cos 30 = 34.641 m/s. The vertical component of the initial...

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The marble is shot at an angle of 30 degrees to the horizontal and has an initial velocity of 40 m/s. The initial velocity of the marble can be divided into 2 components. The horizontal component is equal to 40*cos 30 = 34.641 m/s. The vertical component of the initial velocity is equal to 40*sin 30 = 20 m/s.

When the ball is launched the vertical velocity decreases at 9.8 m/s^2 which is the acceleration due to gravity. If the time taken for the velocity to become 0 is t, 0 = 20 - 9.8*t or t = 2.04 seconds. This is the instant when the ball is at the highest point of its trajectory. As an equal duration of time is taken by the ball to return to the ground the total time taken is 4.08 s.

The horizontal distance traveled by the ball is 34.641*4.08 = 141.33 m

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