A ball is dropped from a height of 80m. The time, in seconds, it takes to reach the ground is:
neglecting air resistance,
h=1/2 g t^2 or time= sqrt (2*h/g)=sqrt (160/9.8) or about four seconds
distance = 1/2*g*t^2
To calculate the time it takes for a ball to fall from a certain height, we can use the equation for freely falling objects:
h = (1/2) * g * t^2
where:
h is the initial height (80m in this case),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time taken to fall.
Rearranging the equation to solve for t:
t = sqrt((2 * h) / g)
Substituting the given values into the equation:
t = sqrt((2 * 80) / 9.8) ≈ sqrt(16.3265) ≈ 4.04 seconds
Therefore, it takes approximately 4.04 seconds for the ball to reach the ground.
To solve this problem, we can use the equation of motion for an object in freefall:
s = ut + (1/2)gt^2
Where:
s = distance (in this case, the height of the ball)
u = initial velocity (which is 0 because the ball is dropped)
g = acceleration due to gravity (which is approximately 9.8 m/s^2)
t = time
Since the ball is dropped from rest, u = 0. Substituting the given values into the equation:
80 = 0 + (1/2)(9.8)t^2
Further simplifying the equation:
80 = 4.9t^2
Divide both sides of the equation by 4.9:
t^2 = 80/4.9
t^2 ≈ 16.33
Now, take the square root of both sides of the equation to solve for t:
t ≈ √16.33
Calculating the square root (√) of 16.33, we find:
t ≈ 4.04 (rounded to two decimal places)
Therefore, it takes approximately 4.04 seconds for the ball to reach the ground when dropped from a height of 80m.