Question

A body of 2 kg mass is dropped from rest, from a height of 10 m. Calculate, after how much time will it strike the earth surface. (g=10m/s square)

Answer 1

Answer:

• Mass of the body (m) = 2 kg
• Height from which body is dropped (H) = -10 m
• Acceleration due to gravity (g) = -10 m/s²
• Initial velocity (u) = 0 m/s
• Time elapsed (t) = ?

By using second kinematical equation of motion we have :

Method (1) :-

=> -H = ut + ½ (-g)t²

=> -10 = 0(t) + ½ × (-10)t²

=> -10 = ½ × - 10 × t²

=> -10 = -5 × t²

=> t² = -10 ÷ -5

=> t² = 10 ÷ 5

=> t² = 2

=> t = √2

=> t = 1.41 seconds

Method (2) :-

=> -H = ut + ½ (-g)t²

=> -H = 0 + ½ (-g)t²

=> -H = ½ × -g × t²

=> H = ½ × g × t²

=> 2H = g × t²

=> t² = 2H/g

=> t= √(2H/g)

=> t = √[(2 × 10)/(10)]

=> t = √(20/10)

=> t = √2

=> t = 1.41 seconds

Simply you can apply direct formula here which is given below :

• t = √(2H/g)