The body is thrown vertically upwards from a height of 40 m with an initial speed of 5 m / s.

The body is thrown vertically upwards from a height of 40 m with an initial speed of 5 m / s. At what height will the body be in 2 m.

Problem data: h0 (the height from which the given body was thrown upwards) = 40 m; Vн (initial speed) = 5 m / s; t (considered flight time) = 2 s.

Constants: by condition g (acceleration of gravity) = 10 m / s2.

1 way) Duration of ascent: V = 0 = Vн – g * t1, whence we express: t1 = Vн / g = 5/10 = 0.5 s.

Lift height: h1 = Vн * t1 – g * t1 ^ 2/2 = 5 * 0.5 – 10 * 0.5 ^ 2/2 = 1.25 m.

Overall height: h = h0 + h1 = 40 + 1.25 = 41.25 m

Remaining time: t2 = t – 1 = 2 – 0.5 = 1.5 s.

The path in 1.5 s fall: S = g * t2 ^ 2/2 = 10 * 1.5 ^ 2/2 = 11.25 m.

Body height in 2 s: hk = h – S = 41.25 – 11.25 = 30 m.

Method 2) hk = h0 + Vn * t – g * t ^ 2/2 = 40 + 5 * 2 – 10 * 2 ^ 2/2 = 40 + 10 – 20 = 30 m.

Answer: The body will be at a height of 30 m.