Question

A rock is thrown downward from an unknown height above the ground with an initial speed of 6.1 m/s. It strikes the ground 1.7 s later. Determine the initial height of the rock above the ground. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m

Answer 1

The initial height from which a rock is thrown downward with an initial speed of 6.1 m/s and striking the ground after 1.7 s is approximately 24.47 m above ground level.

Step-by-step explanation:

To determine the initial height from which a rock is thrown downward with an initial speed of 6.1 m/s and striking the ground after 1.7 s, we can use the kinematic equation for uniformly accelerated motion:

h = vi ⋅ t + (1/2) ⋅ g ⋅ t^2

Here, h is the height, vi is the initial velocity (6.1 m/s downward), g is the acceleration due to gravity (9.8 m/s2), and t is the time (1.7 s). Plugging in the values:

h = 6.1 m/s ⋅ 1.7 s + (1/2) ⋅ 9.8 m/s2 ⋅ (1.7 s)2

After calculating, we get:

h = 10.37 m + 14.1 m

So, the initial height from which the rock was thrown is approximately 24.47 m above the ground.