A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fired into the air by a spring in the engine's smokestack. The marble, which continued to move with the same forward speed as the engine, rejoined the engine 1 sec after it was fired. The measure of the angle the marble's path made with the horizontal was 61degrees. Use the information to find how high the marble went and how fast the engine was moving.

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azfarali azfarali · Answer 1 · 2020-02-13T07:52:42+01:00

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.

So with y(1) = v - 4.9 = 0 we have

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now, the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

andromache andromache · Answer 2 · 2022-02-08T16:15:14+01:00

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

The vertical component of the marble's flight path relative to the train

should be [tex]y(t) = v\times t - (4.9)\times t^2,[/tex]

Here v is the initial upward velocity

So with y(1) = v - 4.9 = 0 we have

v = 4.9 m/s.

Now the height should be

= y(0.5)

[tex]= (4.9)\times (0.5) - (4.9)\times (0.5)^2 \\\\= (4.9)\times (0.25)[/tex]

= 1.225 m.

Now the velocity is

[tex]V = 4.9 \div tan(61)[/tex]

= 2.716 m/s

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