Two stones are thrown vertically upward from the ground , one with 3 times the initial speed of the other. assume free fall.
a) if the faster stone takes 12.0s to return to the ground, how long will take the slower stone to return? ?
b) if the slower stone reach a max height of h, how high (in term of h) will the faster stone go? ?

The time taken by the slower stone is and the maximum height is

Further explanation:

The formula for time can be expressed as follows,

Given:

The speed of first stone is 3 times the initial speed of the other.

Explanation:

Part (a),

The final velocity of the stones is zero as they are thrown vertically upward.

Let us consider the speed of first stone is 3v.

The speed of the second stone is v.

The time taken by the first stone to reach the ground is

The time taken by the second stone to reach the ground is

The ratio of and can be obtained as follows,

Substitute 12 for in above equation to obtain the value of

Part (b)

The second equation of motion is

The maximumheight of slower stone can be expressed as follows,

The maximum height of faster stone can be expressed as follows,

The ratio of and can be obtained as follow,

The time taken by the slower stone is and the maximum height is

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Speed

Keywords: Two stones, thrown vertically upward, speed, initial speed, 3 times, ground, faster stone, 12 second, return ground, slower stone, max height, H, faster stone, high.

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23