Calculate the maximum height of a ball of mass 1.2kg will attain if projected vertically upward with an initial velocity of 17m/s?

To calculate the maximum height attained by a ball projected vertically upward, we can use the equations of motion. First, we need to find the time it takes for the ball to reach its maximum height.

The initial velocity of the ball is 17 m/s, and we know that the final velocity at the highest point is zero (since the ball momentarily stops before falling back down). Using the equation of motion:

v = u + at

Where:
v = final velocity (which is 0 in this case)
u = initial velocity
a = acceleration
t = time

Rearranging the equation gives:

t = (v - u) / a

Since the ball is moving vertically upward, the acceleration due to gravity acts in the opposite direction, which is -9.8 m/s². Plugging in the values into the equation, we have:

t = (0 - 17) / (-9.8)
t = 17 / 9.8
t ≈ 1.7347 seconds

Now, to find the maximum height (h), we can use the equation:

h = ut + (1/2)at²

Where:
h = maximum height
u = initial velocity
t = time (which we just found)
a = acceleration

Plugging in the values:

h = 17 * 1.7347 + (1/2)(-9.8)(1.7347)²
h ≈ 14.8 meters

Therefore, the maximum height attained by the ball is approximately 14.8 meters.