A stone was released from a height of 50 m above the surface of the Earth

A stone was released from a height of 50 m above the surface of the Earth without initial velocity. At what height will it be after 2 seconds of fall?

Data: h0 (the height from which the stone was released) = 50 m; V0 (initial speed) = 0 m / s; t (considered fall time) = 2 s.

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

To calculate the height at which the thrown stone will be in 2 s, we apply the formula: h = h0 – h1 = h0 – V0 * t – a * t ^ 2/2 = h0 – 0 – g * t ^ 2/2.

Let’s perform the calculation: h = 50 – 10 * 2 ^ 2/2 = 50 – 20 = 30 m = 30,000 mm.

Answer: After 2 seconds of falling, the taken stone will be at a height of 30,000 mm.