A Ball Thrown Vertically Upwards From Ground Level Hit The Ground After 4s Calculate The Maxium Height During This Height
40m\s
To calculate the maximum height reached by the ball, we need to understand the motion of the ball when thrown vertically upwards.
When a ball is thrown vertically upwards, it undergoes free fall due to the force of gravity. The motion consists of two phases: the upward phase and the downward phase.
During the upward phase, the initial velocity of the ball decreases uniformly until it reaches zero at the highest point. At this point, the ball changes direction and begins to fall downwards.
During the downward phase, the ball accelerates due to gravity until it hits the ground.
Given that the ball hits the ground after 4 seconds, we can assume that it takes 2 seconds to reach the maximum height. This is because the time taken for the upward and downward phases of motion is equal.
Now, to calculate the maximum height, we need to use the equations of motion. The key equation for this scenario is:
v = u + at
Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2 for objects near the earth's surface)
t = time (2 seconds for the upward phase)
Rearranging the equation, we have:
u = v - at
Substituting the values, we get:
u = 0 - (-9.8 m/s^2 * 2 s)
u = 19.6 m/s
So, the initial upward velocity of the ball is 19.6 m/s.
To calculate the maximum height, we can use the formula:
s = ut + (1/2)at^2
Where:
s = displacement (maximum height)
u = initial velocity (19.6 m/s)
t = time (2 seconds)
a = acceleration due to gravity (-9.8 m/s^2)
Substituting the values, we get:
s = (19.6 m/s * 2 s) + (1/2)(-9.8 m/s^2)(2 s)^2
s = 39.2 m - 19.6 m
s = 19.6 m
Therefore, the maximum height reached by the ball is 19.6 meters.