A 49-kg lead ball is dropped from the leaning tower of Pisa. The tower is 55 m high.
(c) What is the speed of the ball 1.3 s after it is released?
acceleration is 9.8m/s^2
That is, every second in increases by 9.8m/s
So, v after 1.3 seconds = 9.8*1.3 = 12.74m/s
99m
To find the speed of the ball 1.3 seconds after it is released, we can use the equations of motion.
First, we need to determine the time it takes for the ball to fall from the top of the tower to the point 1.3 seconds later.
The equation for the height of a falling object as a function of time is:
h = 0.5 * g * t^2
Where:
h = height
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
In this case, we need to find the time it takes for the object to fall from a height of 55 meters to a height of h1 = 55 meters - h2 = 0.5 * g * (1.3 seconds)^2 = 9.735 meters.
Using the equation for height, we can rearrange it to solve for time:
t = sqrt(2 * h / g)
Plugging in the values, we get:
t = sqrt(2 * 9.735 m / 9.8 m/s^2) = 1.506 seconds (approximately)
Now that we have the time it takes for the ball to fall that distance, we can use the equation for the final velocity of a falling object:
v = g * t
Plugging in the value of g (9.8 m/s^2) and t (1.506 seconds), we can calculate the velocity:
v = 9.8 m/s^2 * 1.506 seconds = 14.7948 m/s (approximately)
Therefore, the speed of the ball 1.3 seconds after it is released is approximately 14.79 m/s.