Question

A rock is tossed straight up with a speed of 30 m/s. When it returns, it falls into a hole 14 m deep. What is the maximum height reached by the rock?

1) 14 m
2) 30 m
3) 44 m
4) Cannot be determined

Answer 1

The maximum height reached by the rock is 44 m.

Step-by-step explanation:

To find the maximum height reached by the rock, we can use the equation
`v^2 = u^2 - 2as`, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

In this case, the final velocity is 0 m/s since the rock reaches its maximum height. The initial velocity is 30 m/s and the acceleration due to gravity is
`9.8 m/s^2`. The displacement is the sum of the depth of the hole (14 m) and the maximum height reached by the rock.

Plugging these values into the equation, we can solve for the maximum height:

`0^2 = 30^2 - 2 * 9.8 * (14 + h)`

Simplifying the equation, we get:

-882 = -274.4 - 19.6h

h = (882 + 274.4) / 19.6

= 44 m

Therefore, the maximum height reached by the rock is 44 m.