An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s²
We can list the data as,
height covered or distance covered, s = 500m
g = 10m/s²
To find - (a) initial velocity u = ?
(b) time taken by object , t = ?
Solution -
Using Newton’s third equation of motion
v² = u² +2as
putting the values , v =0m/s , a = -g , s = 500m
0 = u² +2 x (-g) x 500
0 = u² -2 x g x 500
u² = 2 x g x 500 = 1000g = 1000 x 10= 10000
taking square root on both sides, we get
u = 100m/s
To find the time taken t,
t = time taken to go from bottom to top + time taken to go from top to bottom
t = t₁ + t₂
To find t₁
using Newton’s first equation
v = u + at
initial velocity, u = 100m/s
final velocity, v= 0m/s
a = -g = -10m/s²
t= t₁
0 = 100 - 10 x t₁
10t₁ = 100
t₁ = 10s
To find t₂
As the object comes down back on the ground,
According to Newton’s second equation
