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Daniel [21]
2 years ago
6

A baseball is dropped off the side of a cliff. It free-falls to the ground in 8 seconds. During the third second, the ball is tr

aveling about ___ m/s.
Physics
2 answers:
Anna [14]2 years ago
6 0

when an object drops it accelerates at 9.8 mps so at 3 seconds the equation would be 9.8x3 which equals 29.4

kozerog [31]2 years ago
3 0
The acceleration of gravity is 9.8 m/s² .  That means that a falling object
is always falling 9.8 m/s faster than it was falling 1 second earlier.

If an object is not slowed by air resistance, and has far enough to go
so that it's still falling after three whole seconds, then at the end of
three seconds it's falling at

           (9.8 m/s²) x (3 sec)  =  29.4 m/s
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A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
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Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

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