Question

An object of mass 10 kg is thrown vertically upwards with the velocity of 20 m/s. Find the maximum height reach the object (g = 10 m/s2)
a)20m
b) 5m
c) 10m
d) 40m

Answer 1

Answer:

a) 20 m

Explanation:

height  = (initial velocity)/2 x g

therefore 20^2/2 x 10

= 20 m

Answer 2

Explanation:

Mass of the object = 10kg

Initial velocity (u)= 20m/s

Final velocity (v) = 0m/s

As the stone is thrown upward the acceleration due to gravity is to be taken negative.

g = acceleration due to gravity (a) = -10m/s²

Using 1st equation of motion:-

→ v = u + at

→ 0 = 20 + (-10)t

→ -20 = -10t

→ t = -20/-10 = 2

Time taken (t) = 2sec

Using 3rd equation of motion:-

→ s = ut + ½ at²

→ s = (20 × 2) + ½ × (-10) × (2)²

→ s = 40+ (-5) × 4

→ s = 40 - 20

→ s = 20m

∴ Maximum height reached by the object = 20m

Alternatively:-

By using 2nd equation of motion:-

→ v² - u² = 2as

→ (0)² - (20)² = 2 × (-10) × s

→ - 400 = -20s

→ s = -400/-20

→ s = 20m

∴ Maximum height reached by the object = 20m