To calculate how high the ball goes, we can use the equations of motion for vertical motion under constant acceleration due to gravity. The key equation to use in this case is:
v = u + at
where:
v = final velocity (in this case, the ball's velocity at its highest point, which is 0 m/s since the ball momentarily comes to rest at its peak)
u = initial velocity (20 m/s, given in the problem)
a = acceleration due to gravity (-9.8 m/s^2, assuming downward as positive as per convention)
t = time taken to reach the highest point (what we're trying to find)
We can rearrange the equation to solve for time:
t = (v - u) / a
Plugging in the known values:
v = 0 m/s
u = 20 m/s
a = -9.8 m/s^2
t = (0 - 20) / -9.8 = 20 / 9.8 ≈ 2.04 seconds (rounded to two decimal places)
So, the ball takes approximately 2.04 seconds to reach its highest point.
Now, we can use another equation of motion to find the height the ball reaches at its highest point:
s = ut + 0.5at^2
where:
s = height (distance) traveled by the ball (what we're trying to find)
u = initial velocity (20 m/s)
t = time taken to reach the highest point (2.04 seconds, as calculated above)
a = acceleration due to gravity (-9.8 m/s^2)
Plugging in the known values:
u = 20 m/s
t = 2.04 seconds
a = -9.8 m/s^2
s = 20 * 2.04 + 0.5 * (-9.8) * (2.04)^2 ≈ 20.4 meters (rounded to one decimal place)
So, the ball reaches a height of approximately 20.4 meters above its initial position before coming to a stop momentarily and then falling back down due to gravity.