An arrow fired from a bow rose to a height of 30 m. What was the speed of the arrow at a height of 10 m?

h = 30 m.

V = 0 m / s.

h1 = 10 m.

g = 9.8 m / s ^ 2.

V1 -?

Only one gravity force m * g, directed vertically downward, acts on an arrow fired from a bow. The force of gravity is constant in magnitude and direction, therefore, according to 2 Newton’s laws, the arrow will move with constant acceleration – free fall g.

Since the acceleration of gravity g is directed in the opposite direction of the movement of the arrow, it will be decelerated.

Boom initial speed V0
For uniformly accelerated motion with the acceleration of the free fall of the arrow upward, the following kinematic formulas are valid:

h = V0 * t – g * t ^ 2/2;
h = (V0 ^ 2 – V ^ 2) / 2 * g;
g = (V0 – V) / t.
Where V, V0 – final and initial boom speed, t – boom lifting time, h – lifting height.

Since the final speed of the arrow V at a height of h = 30 m is equal to zero V = 0, the arrow has stopped, the formulas will take the form:

h = V0 * t – g * t ^ 2/2;
h = V0 ^ 2/2 * g;
g = V0 / t.
Let us express the square of the initial speed of the arrow from the formula: h = V0 ^ 2/2 * g.

V0 ^ 2 = 2 * g * h.

Body speed at height h1
Let’s use the second formula to find the speed of the arrow at the height h1: h1 = (V0 ^ 2 – V1 ^ 2) / 2 * g.

2 * g * h1 = V0 ^ 2 – V1 ^ 2;

V1 ^ 2 = V0 ^ 2 – 2 * g * h1;

V1 ^ 2 = 2 * g * h – 2 * g * h1;

V1 ^ 2 = 2 * g * (h – h1);

The formula for determining the boom speed V1 at the height h1 will be: V1 = √ (2 * g * (h – h1)).

V1 = √ (2 * 9.8 m / s ^ 2 * (30 m – 10 m)) = 19.8 m / s.

Answer: the speed of the boom at a height of h1 = 10 m is V1 = 19.8 m / s.