An object is thrown upwards at an initial speed of 19.6 m/s from a height of 65m. Determine how long will take for the object to reach the ground (time of flight). what formula should i use?
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To determine the time it takes for the object to reach the ground (time of flight), we can use the kinematic equation:
h = ut + (1/2)at^2
Where:
h = initial height (65m)
u = initial velocity (19.6 m/s)
a = acceleration due to gravity (-9.8 m/s^2, considering it is acting in the downward direction)
t = time
Since the object is thrown upwards, the final height will be zero when it reaches the ground. Plugging in the known values into the equation, we get:
0 = (19.6)t + (0.5)(-9.8)(t^2)
Simplifying the equation gives:
0 = 19.6t - 4.9t^2
Rearranging the equation, we have:
4.9t^2 - 19.6t = 0
Now, we can factor out t from the equation:
t(4.9t - 19.6) = 0
This equation has two solutions: t = 0 and 4.9t - 19.6 = 0.
The first solution (t = 0) represents the initial time when the object is thrown, so we need to consider the second solution:
4.9t - 19.6 = 0
Solving for t, we get:
4.9t = 19.6
t = 19.6 / 4.9
t = 4 seconds
Therefore, it will take 4 seconds for the object to reach the ground (time of flight).