an object is launched vertically in the air at 29.4 meters per second from 10 meter tall platform . find its maximum height in seconds and in meters ?
To find the maximum height reached by the object, we need to calculate the time it takes for the object to reach its peak height.
The motion of the object can be described using the equations of motion. In this case, we can use the equation for vertical displacement:
s = ut + (1/2)at^2
Where:
s = vertical displacement (maximum height)
u = initial velocity (29.4 m/s)
t = time taken
a = acceleration (acceleration due to gravity, which is approximately -9.8 m/s^2)
Considering that the object is launched vertically upwards, the final velocity at the maximum height is 0 m/s. Therefore, we can use the equation:
0 = 29.4 + (-9.8)t
Simplifying the equation, we have:
-29.4 = -9.8t
Dividing both sides by -9.8, we find:
t = 3 seconds
Now that we know the time it takes to reach the maximum height is 3 seconds, we can calculate the vertical displacement using the equation:
s = ut + (1/2)at^2
s = (29.4)(3) + (1/2)(-9.8)(3)^2
s = 88.2 - 44.1
s = 44.1 meters
Therefore, the maximum height reached by the object is 44.1 meters, and it reaches this height after 3 seconds.