A stone thrown at an angle of 45o to the horizon, 0.8 s after the start of movement, continued to rise

A stone thrown at an angle of 45o to the horizon, 0.8 s after the start of movement, continued to rise and had a vertical velocity component of 12 m / s. What is the distance between the place of throwing and the place where the stone falls?

t = 0.8 s.

Vу = 12 m / s.

∠α = 45 °.

g = 10 m / s2.

L -?

Since only gravity, directed vertically downward, acts on the stone during movement, the movement of the stone can be divided into two types: horizontally, it moves uniformly with a speed Vx = V0 * sinα, vertically uniformly accelerated, with gravitational acceleration g and an initial vertical speed Vо = V0 * cosα.

The vertical component of the velocity Vу will change according to the law: Vу = Vу – g * t = V0 * cosα – g * t.

V0 * cosα = Vу + g * t.

V0 = (Vу + g * t) / cosα.

V0 = (12 m / s + 10 m / s2 * 0.8 s) / cos45 ° = 28.3 m / s.

The flight range L is expressed by the formula: L = V0 * cosα * t1, where V0 is the speed of the grenade throw, t1 is the flight time of the grenade.

g = V0 * sinα / t2, where t2 is the time for the grenade to rise to the maximum height.

t2 = V0 * sinα / g.

t1 = 2 * t ^ 2 = 2 * V0 * sinα / g.

L = V0 * cosα * 2 * V0 * sinα / g = V0 ^ 2 * sin2α / g.

L = (28.3 m / s) ^ 2 * sin2 * 45 ° / 10 m / s2 = 80 m.

Answer: the flight range of the stone was L = 80 m.