Question

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.


A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?​

Answer 1

Explanation:

IN QUESTION 1

Given data:

Initial velocity

u = 0

Tower height = total distance = 19.6m

g = 9.8 m/s2

Consider third equation of motion

v2 = u2 + 2gs

v2 = 0 + 2 × 9.8 × 19.6

v2 = 384.16

v = √(384.16)

v = 19.6m/s

IN QUESTION 2

Given data:

Initial velocity u = 40m/s

g = 10 m/s2

Max height final velocity = 0

Consider third equation of motion

v2 = u2 – 2gs [negative as the object goes up]

0 = (40)2 – 2 x 10 x s

s = (40 x 40) / 20

Maximum height s = 80m

Total Distance = s + s = 80 + 80

Total Distance = 160m

Total displacement = 0 (The first point is the same as the last point

hope it helps you

Answer 2

Explanation:

hope you get it

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