Question

A stone at a height of 5 m is thrown horizontally and falls on the ground at a horizontal distance of10 m away from where it was thrown. What is the velocity of stone at the point of impact with ground

Answer 1

Answer:

The velocity of stone at the point of impact with ground is 14 m/s.

Solution:

Along x component,

» ux = 0 m/s, sx = 5 m, ax = 9.8 m/s²

By Newton's third law of motion

» v²x = u²x + 2 (ax) (sx)

» v²x = 0 + 2 (9.8) (5)

» v²x = 2 (9.8) (5)

» v = 9.9 m/s (approx)

By Newton's first law of motion

» vx = ux + (ax) t

» 9.9 = 0 + (9.8) t

» 9.9 = 0 + (9.8) t» t = 1.01 s

Along y component,

Acceleration is zero.

=> Speed = Distance/Time

=> vy = 10/1.01 = 9.9 m/s

Hence, v = √(v²x + v²y)

» v = √(9.9² + 9.9²)

» v = 9.9 × √2

» v = 9.9 × 1.41

» v = 14 m/s (approx)