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A dart gun is fired horizontally from a height of 1.5 m above the floor. The dart then lands on the floor

4 meters from the edge of the table. Determine the initial velocity of the marble as it leaves the
launcher.

1 Answer

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Let v be the speed of the projectile, be it a dart or a marble. It covers a horizontal distance x with height y at time t according to

x = v t

y = 1.5 m - 1/2 g t²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The projectile lands 4 m from the table, so it hits the ground at time t such that

4 m = v t ==> t = (4 m) / v

It's on the ground, so that y = 0 at this time. Substitute this into the height equation and solve for v :

0 = 1.5 m - 1/2 g ((4 m) / v

v² = (g (16 m²)) / (3.0 m)

v ≈ 7.23 m/s

User Dave Collins
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