A ball thrown horizontally from a height of 2m above the ground fell at a distance of 7m. Find the start and end speed.

h = 2 m.

g = 10 m / s2.

L = 7 m.

V0 -?

V -?

The movement of the ball can be divided into two types: horizontally, it moves uniformly with a speed Vx = V0, vertically uniformly accelerated with the acceleration of gravity g.

The flight range L is expressed by the formula: L = V0 * t, where V0 is the initial speed of the ball, t is the flight time of the ball.

V0 = L / t.

Let’s find the time of flight of the ball t from the vertical motion.

h = g * t ^ 2/2.

t = √ (2 * h / g).

t = √ (2 * 2 m / 10 m / s2) = 0.63 s.

V0 = 7 m / 0.63 s = 11.1 m / s.

Let’s find the vertical speed of the ball Vх at the moment of hitting the ground: Vу = g * t.

Vу = 10 m / s2 * 0.63 s = 6.3 m / s.

At the moment of hitting the ground, the ball will have a horizontal component of velocity Vх and a vertical component Vу.

Therefore, V = √ (Vx ^ 2 + Vy ^ 2).

V = √ ((11.1 m / s) ^ 2 + (6.3 m / s) ^ 2) = 12.76 m / s.

Answer: V0 = 11.1 m / s, V = 12.76 m / s.