an object falls from a 3m height what is the speed of the object just before it touches the ground?
The object experiences uniformly accelerated motion (UAM) so we can use the formula:
vf^2 - vo^2 = 2gd
where
vf = final velocity
vo = initial velocity
g = acceleration due to gravity = 9.8 m/s^2
d = distance
Since the object is in free fall, vo is zero:
vf^2 - 0 = 2*9.8*3
vf^2 = 58.8
vf = 7.67 m/s
hope this helps~ `u`
To find the speed of an object just before it touches the ground when it falls from a given height, we can use the equation for free fall motion:
v^2 = u^2 + 2as
Where:
v is the final velocity (speed of the object just before it touches the ground)
u is the initial velocity (which is 0 in this case as the object is dropped)
a is the acceleration due to gravity (approximately 9.8 m/s^2)
s is the distance or height through which the object falls
In this case, the height (s) is given as 3 meters.
Step 1: Substitute the values into the equation.
v^2 = 0^2 + 2 * 9.8 * 3
Step 2: Simplify the equation.
v^2 = 58.8
Step 3: Take the square root of both sides of the equation to find the speed (v).
v = √58.8
Using a calculator, the speed (v) is approximately 7.67 m/s just before the object touches the ground.
To calculate the speed of an object just before it touches the ground when falling from a certain height, you can use the equation of motion for free fall:
v^2 = u^2 + 2as
Where:
v is the final velocity
u is the initial velocity (usually taken as 0 for objects in free fall)
a is the acceleration due to gravity (approximately 9.8 m/s^2)
s is the distance fallen
In this case, the distance fallen is the height of 3 meters. However, since the object is falling, we take the negative direction as the downward direction, so s will be -3 meters.
Plugging these values into the equation:
v^2 = 0^2 + 2 * 9.8 * (-3)
Simplifying the equation:
v^2 = -58.8
Taking the square root of both sides:
v = √(-58.8)
Since we are dealing with physical quantities, we must consider only positive values, so we take the positive square root:
v = √58.8
Therefore, the speed of the object just before it touches the ground is approximately 7.67 m/s (rounded to two decimal places).