One second after being thrown straight down an object is falling with a speed of 20m/s. How fast will it be falling 2 seconds later
To determine the speed of the falling object 2 seconds later, we can use the following equation:
v = u + at
Where:
v = final velocity (speed) of the object
u = initial velocity (speed) of the object
a = acceleration of the object
t = time
Given information:
u = 20 m/s
t = 2 seconds
Since the object is falling straight down, we can assume the acceleration due to gravity is acting on it. Therefore, acceleration (a) is equivalent to the acceleration due to gravity, which is approximately 9.8 m/s².
Let's substitute the given values into the equation:
v = u + at
v = 20 m/s + (9.8 m/s²)(2 s)
Calculating the equation:
v = 20 m/s + 19.6 m/s
v = 39.6 m/s
Therefore, the falling object will be moving at a speed of 39.6 m/s 2 seconds later.
To determine the speed of an object falling 2 seconds later, we can use the equations of motion for uniformly accelerated linear motion. The equation that relates the final velocity (v), initial velocity (u), acceleration (a), and time (t) is:
v = u + at
In this case, the initial velocity (u) is 20 m/s, the acceleration due to gravity (a) is approximately 9.8 m/s² (assuming no other external forces acting on the object), and the time (t) is 2 seconds. We want to find the final velocity (v).
Plugging in the values, the equation becomes:
v = 20 m/s + (9.8 m/s²)(2 s)
Calculating the right-hand side:
v = 20 m/s + 19.6 m/s
v = 39.6 m/s
Therefore, 2 seconds later, the object will be falling with a speed of 39.6 m/s.
g=9.8 (m/s) /s
After two further seconds in a free-fall without air resistance, the increase in velocity is
δv=g(2 s)=9.8*2 m/s
=19.6 s
which allows you to figure out the final velocity.