Question

As preparation for this problem, review Conceptual Example 9. From the top of a cliff overlooking a lake, a person throws two stones, as shown in the drawing. The cliff is 35.0 m high. The two stones described have identical initial speeds of v₀ = 17.6 m/s and are thrown at an angle θ = 31.3 °, one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the water? Neglect air resistance.

Answer 1

Given:

Initial launch height for both stones,

Initial launch velocity of both stones,

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Consider the stone thrown at an angle

Since the stone is thrown at an angle below the horizontal, initial vertical velocity component, is acting downwards.

First, find the time of flight

Consider the vertical motion of the stone

Use formula

---------Points to consider:

1. When the stone strikes the water, becomes zero

2. Initial vertical velocity component, of the stone and is acting downwards so put negative sign for both.

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Solve the quadratic equation with a calculator

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Consider the horizontal motion of the stone

There is no acceleration in the horizontal direction, so horizontal velocity remains constant.

Horizontal velocity = Horizontal distance/time

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Consider the stone thrown at an angle

Since the stone is thrown at an angle above the horizontal, initial vertical velocity component, is acting upwards.

First, find the time of flight

Consider the vertical motion of the stone

Use formula

---------Points to consider:

1. When the stone strikes the water, becomes zero

2. is acting downwards so put a negative sign.

--------------

Solve the quadratic equation with a calculator

--------------

Consider the horizontal motion of the stone

There is no acceleration in the horizontal direction, so horizontal velocity remains constant.

Horizontal velocity = Horizontal distance/time

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Distance between the points where the stones strike,

ANSWER:

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