Question

A stone is dropped from the top of a tower which touches the ground after 2 seconds. Calculate the speed of the stone hitting the ground surface. (g = 9.8 m/s²)​

Answer 1

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`\mathbb{ QUESTION:}`

A stone is dropped from the top of a tower which touches the ground after 2 seconds.

• Calculate the speed of the stone hitting the ground surface. (g = 9.8 m/s²).

`\mathbb{SOLUTION:}`

• 
  `v = u + at`

• 
  `v = 9.8 m/s² x 2sec`

• 
  `v = 19.6 \: m/ {s}^(2)`

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Hence, The stone hit the ground with the velocity speed of 19m/s².

- When you're solving velocity, you're determining how fast an object moves from its original position, with respect to a frame of reference, and a function of time.

- That means an object's velocity will be equal to the object's speed and direction of motion.

"Problem has been solve"

(ノ^_^)ノ

Answer 2

19.6 m/s

Step-by-step explanation:

The velocity is the product of acceleration and time.

__

Here, the acceleration is due to gravity.

v = at

v = (9.8 m/s²)(2 s) = 19.6 m/s

The stone hits the ground with a speed of 19.6 m/s.