a ball thrown vertically upward from the ground,hit the ground after 4seconds,calculate the maximum height reach by the ball during the jorney
time up equals time down
h = 1/2 g t^2 = 4.9 m/s^2 * (2 s)^2
I don't know pls can u fully explain the answer
To calculate the maximum height reached by the ball, we can use the equations of motion. When the ball reaches its maximum height, its vertical velocity becomes zero. We can use the following equation:
v = u + at
Where:
- v is the final velocity (0 m/s at the maximum height)
- u is the initial velocity (when the ball is thrown upwards)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken to reach the maximum height
Given:
- The time taken to reach the ground is 4 seconds. This is the total time for the ball’s journey, so it took 2 seconds for the ball to reach the maximum height.
Using the equation above, we can calculate the initial velocity (u) of the ball:
0 = u - 9.8 * 2
Rearranging the equation, we have:
u = 9.8 * 2
u = 19.6 m/s
Now, we can calculate the maximum height (h):
h = ut + (1/2)at^2
Where:
- h is the maximum height
Using the above equation:
h = 19.6 * 2 + (1/2)(-9.8)(2)^2
h = 39.2 + (-4.9)(4)
h = 39.2 - 19.6
h = 19.6 meters
Therefore, the maximum height reached by the ball during its journey is 19.6 meters.
To calculate the maximum height reached by the ball, we can use the equation for the motion of a vertically thrown object.
The equation for the motion is:
h = v0*t - (1/2)g*t^2
Where:
- h is the height of the object
- v0 is the initial velocity
- t is the time
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Given that the ball hits the ground after 4 seconds, we can assume that the maximum height occurs halfway through its journey, which is at 2 seconds.
Now, we need to find the initial velocity (v0).
Since the ball is thrown vertically upward, it will reach its highest point when its vertical velocity becomes zero before starting to fall. At that point, the only force acting on the ball is the acceleration due to gravity. Therefore, the final velocity at the highest point will be zero.
Using the equation for velocity:
v = v0 - g*t
At t = 2 seconds, v = 0, and g = 9.8 m/s^2. Plugging these values into the equation and solving for v0:
0 = v0 - 9.8*2
v0 = 19.6 m/s
Now that we have the initial velocity (v0), we can calculate the maximum height (h) using the equation:
h = v0*t - (1/2)g*t^2
At t = 2 seconds:
h = 19.6*2 - (1/2)*9.8*(2^2)
h = 39.2 - 19.6
h = 19.6 m
Therefore, the maximum height reached by the ball during its journey is 19.6 meters.