Both the feather and the ball will reach the ground at the same time in a vacuum due to the acceleration due to gravity being constant for all objects. However, in the presence of air resistance, the feather will take longer to reach the ground compared to the ball due to its larger surface area and lighter weight.
The ball will take approximately 1.74 seconds to reach the ground. This can be calculated using the equation ( t = \sqrt{\frac{2h}{g}} ), where ( h = 12.0 , m ) and ( g = 9.81 , m/s^2 ).
Using the equation ( h = 0.5 \times g \times t^2 ) where ( h = 1.4 , m ) and ( g = 9.81 , m/s^2 ), we can solve for ( t ). Plugging in the values, we get ( 1.4 = 0.5 \times 9.81 \times t^2 ). Solving for ( t ), we find that it takes approximately 0.53 seconds for the ball to reach the ground.
The time it takes for the ball to reach the ground can be calculated using the equation: t = sqrt(2h/g), where: t = time h = initial height (18m) g = gravitational acceleration (36 m/s^2) Plugging in the values, we get: t = sqrt(2*18/36) = sqrt(1/2) = 0.707 seconds.
When an object is dropped from a certain height, the time it takes to reach the ground is independent of the height (assuming no air resistance). Therefore, whether you drop the object from three times the initial height or the original height, it will still take the same time (T) to reach the ground.
that depends on how high it is thrown and whats its atomic mass is.
Both the feather and the ball will reach the ground at the same time in a vacuum due to the acceleration due to gravity being constant for all objects. However, in the presence of air resistance, the feather will take longer to reach the ground compared to the ball due to its larger surface area and lighter weight.
The ball will take approximately 1.74 seconds to reach the ground. This can be calculated using the equation ( t = \sqrt{\frac{2h}{g}} ), where ( h = 12.0 , m ) and ( g = 9.81 , m/s^2 ).
t matters how much mass the ball has
"3.2" or "3.20" please put all of that
25 minutes
It takes about four days.
Probably. It has nothing to do with weight. Weight does not play a factor in how fast objects fall, however, size, wind resistance, and aerodynamics do. The bigger an object, the more air hits it going up as the object falls, and therefore slows its descent. Also, if you drop a feather off a building it will take a while to reach the ground, since the little hair-type things sticking out from the feather "catch" the air and slow it down. Then there is aerodynamics. The more aerodynamic an object is, the easier it can cut through the air, therefore, if you drop a very aerodynamic object off a building it will not take as long for it to reach the ground. In this case, we can assume both are pretty much the same shape, however, the beach ball is much larger, and catches the air on the way down. That's what makes it fall so much slower.
Test it like this: Hold the ball as high as your arm will reach without lifting your toes. Then, let it do a fall with no acceleration. If the ball reaches your head with the first bounce it is fully blown up.
Using the equation ( h = 0.5 \times g \times t^2 ) where ( h = 1.4 , m ) and ( g = 9.81 , m/s^2 ), we can solve for ( t ). Plugging in the values, we get ( 1.4 = 0.5 \times 9.81 \times t^2 ). Solving for ( t ), we find that it takes approximately 0.53 seconds for the ball to reach the ground.
yes
The time it takes for the ball to reach the ground can be calculated using the equation: t = sqrt(2h/g), where: t = time h = initial height (18m) g = gravitational acceleration (36 m/s^2) Plugging in the values, we get: t = sqrt(2*18/36) = sqrt(1/2) = 0.707 seconds.