Question

A stone is thrown horizantally with a velocity of 50 m/s from the top of a building. Find its velocity after 2 seconds.

Answer 1

V = 53.7 m/s ∠-21.41°

Explanation:

Since the stone's initial vertical velocity, u is 0m/s(since it is thrown horizontally), we find the stone's final vertical velocity, v from v = u - gt where u and v have the meanings as given above, g = acceleration due to gravity = 9.8 m/s² and t = time = 2 s

So, v = u - gt

v = 0 m/s - 9.8 m/s² × 2 s

v = 0 m/s - 19.6 m/s

v = - 19.6 m/s

Since the stone's horizontal velocity is u' = 50 m/s, we find the resultant magnitude of velocity V from V = √(u'² + v²)

So, V = √((50 m/s)² + (-19.6 m/s)²)

V = √(2500 m²/s² + 384.16 m²/s²)

V = √(2884.16 m²/s²)

V = 53.7 m/s

Its direction Ф = tan⁻¹(v/u')

Ф = tan⁻¹(-19.6 m/s/50 m/s)

Ф = tan⁻¹(-0.392)

Ф = -21.41°

So it velocity V = 53.7 m/s ∠ -21.41°