A child drops a coin from the top of a building which is 50m high. Calculate velocity of the coin and the time taken by the coin to reach the ground
Answers
Given :-
A child drops a coin from the top of a building which is 50 meters high .
Required to find :-
- Velocity of the coin with which it strikes the ground ?
- Time taken by the coin to reach the ground ?
Equations used :-
v² - u² = 2as
s = ut + ½ at²
Solution :-
Given information :-
A child drops a coin from the top of a building which is 50 meters high .
we need to find ;
Velocity of the coin with which it strikes the ground ?
Time taken by the coin to reach the ground ?
So,
From the given data we can conclude that ;
- Initial velocity of the coin ( u ) = 0 m/s
- Height ( s ) = 50 meters
Since, the coin is falling freely . The acceleration due to gravity comes into play .
The value of the g is positive because the movement is in the direction of the acceleration due to gravity .
- Acceleration due to gravity ( g ) = 10 m/s²
Using the equation of motion ;
i.e. v² - u² = 2as
v² - ( 0 )² = 2 x 10 x 50
v² - 0 = 2 x 10 x 50
v² = 20 x 50
v² = 1000
v = √1000
v = 31.6 m/s.
Hence,
- Final velocity ( v ) = 31.6 m/s
Now,
Let's find out the time taken ;
Using the Equation of motion ;
i.e. s = ut + ½ at²
50 = 0 x t + ½ x 10 x t x t
50 = 0 + ½ x 10 x t x t
50 = 0 + ½ x 10 x t²
50 = 0 + 5 x t²
50 = 0 + 5t²
50 = 5t²
0 = 5t² - 50
5t² - 50 = 0
5 ( t² - 10 ) = 0
t² - 10 = 0/5
t² - 10 = 0
t² = 10
t = √10
t = + 3.16 or - 3.16
Since,
Time can't be in negative .
Hence,
- Time taken ( t ) = 3.16 seconds
Therefore,
The velocity at which it strikes the ground = 31.6 m/s
Time taken to fall through this height = 3.16 seconds
Given ,
Initial velocity (u) = 0 m/s
Displacement (s) = 50 m
We know that , the Newton's first equation is given by
(v)² - (u)² = 2as
(v)² - (0)² = 2 × 10 × 50
(v)² = 1000
v = √1000
v = 31.6 m/s
Now , the Newton's first equation of motion is given by
v = u + at
31.6 = 0 + 10t
t = 31.6/10
t = 3.16 sec