a ball of mass 5kg is dropped from a height of 78.4m,the time taken by the ball to reach the ground is?
mass irrelevant
78.4 = (1/2) g t^2
t^2 = 2 * 78.4 / 9.81
S=0.5gt^2
78.5=0.5*10t^2
t=3.96 s
Note to Benson: Nowhere on this planet is the acceleration due to gravity 10m/s^2.
He has a Uranium mine.
To calculate the time taken by the ball to reach the ground, we can use the principles of free fall motion.
We can start by using the equation:
s = ut + (1/2)gt^2
Where:
s = 78.4m (height)
u = initial velocity (0 m/s since the ball is dropped)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken
Since the ball is dropped, the initial velocity is 0 m/s, so the equation simplifies to:
s = (1/2)gt^2
Rearranging the equation, we get:
t^2 = (2s)/g
Substituting the values:
t^2 = (2 * 78.4) / 9.8
t^2 = 158.4 / 9.8
t^2 ≈ 16.16
To find t, we take the square root of both sides:
t ≈ √16.16
t ≈ 4.02 seconds (rounded to two decimal places)
Therefore, the time taken by the ball to reach the ground is approximately 4.02 seconds.