A ball thrown vertically upward from the ground level hits the ground after 4 seconds calculate the maximum height it reached during its journey
Tf = 0.5 * 4 = 2 s. = Fall time,
h = 0.5g*Tf^2.
To calculate the maximum height reached by the ball, we need to make use of the equations of motion.
First, we need to find the initial velocity of the ball when it was thrown upwards. We know that the time taken for the ball to reach the ground is 4 seconds, and the final velocity when it hits the ground is 0 m/s (as it's at rest). The equation we can use is:
v = u + at
Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. Plugging in the given values, we have:
0 = u + (9.8 m/s²)(4 s)
0 = u + 39.2 m/s
u = -39.2 m/s
So the initial velocity of the ball when thrown upwards is -39.2 m/s. The negative sign indicates that the velocity was in the opposite direction of the gravitational acceleration.
Now, to calculate the maximum height reached by the ball, we can use the second equation of motion:
s = ut + (1/2)at²
Where s is the displacement (change in height), u is the initial velocity, t is the time taken, and a is the acceleration. Rearranging the equation to solve for s, we have:
s = ut + (1/2)at²
s = (-39.2 m/s)(4 s) + (1/2)(-9.8 m/s²)(4 s)²
s = -156.8 m - 78.4 m
s = -235.2 m
The negative sign indicates that the displacement is in the opposite direction of the initial velocity. However, in this case, we are interested in the magnitude of the displacement (height), so we disregard the negative sign:
Maximum height reached = |s| = 235.2 m
Therefore, the maximum height reached by the ball during its journey is 235.2 meters.