Answer:
The time taken for the upward motion is 1 second. The same time is taken for the downward motion
It reaches a maximum height of 4.9 meters.
Step-by-step explanation:
The equation of motion is:
![x(t) = -4.9t^{2} + 9.8t](https://tex.z-dn.net/?f=x%28t%29%20%3D%20-4.9t%5E%7B2%7D%20%2B%209.8t)
Since the term which multiplies t squared is negative, the graph is concave down, that is, x increases until the vertex, where it reaches it's maximum height, then it decreases.
Vertex of a quadratic equation:
Quadratic equation in the format ![x(t) = at^{2} + bt + c](https://tex.z-dn.net/?f=x%28t%29%20%3D%20at%5E%7B2%7D%20%2B%20bt%20%2B%20c)
The vertex is the point
, in which
![t_{v} = -\frac{b}{2a}](https://tex.z-dn.net/?f=t_%7Bv%7D%20%3D%20-%5Cfrac%7Bb%7D%7B2a%7D)
In this question:
![x(t) = -4.9t^{2} + 9.8t](https://tex.z-dn.net/?f=x%28t%29%20%3D%20-4.9t%5E%7B2%7D%20%2B%209.8t)
So ![a = -4.9, b = 9.8](https://tex.z-dn.net/?f=a%20%3D%20-4.9%2C%20b%20%3D%209.8)
Vertex:
![t_{v} = -\frac{9.8}{2*(-4.9)} = 1](https://tex.z-dn.net/?f=t_%7Bv%7D%20%3D%20-%5Cfrac%7B9.8%7D%7B2%2A%28-4.9%29%7D%20%3D%201)
The time taken for the upward motion is 1 second.
![x(t_{v}) = x(1) = 9.8*1 - 4.9*(1)^{2} = 4.9](https://tex.z-dn.net/?f=x%28t_%7Bv%7D%29%20%3D%20x%281%29%20%3D%209.8%2A1%20-%204.9%2A%281%29%5E%7B2%7D%20%3D%204.9)
It reaches a maximum height of 4.9 meters.
Downward motion:
From the vertex to the ground.
The ground is t when x = 0. So
![-4.9t^{2} + 9.8t = 0](https://tex.z-dn.net/?f=-4.9t%5E%7B2%7D%20%2B%209.8t%20%3D%200)
![4.9t^{2} - 9.8t = 0](https://tex.z-dn.net/?f=4.9t%5E%7B2%7D%20-%209.8t%20%3D%200)
![4.9t(t - 2) = 0](https://tex.z-dn.net/?f=4.9t%28t%20-%202%29%20%3D%200)
![4.9t = 0](https://tex.z-dn.net/?f=4.9t%20%3D%200)
Or
![t - 2 = 0](https://tex.z-dn.net/?f=t%20-%202%20%3D%200)
![t = 2](https://tex.z-dn.net/?f=t%20%3D%202)
It reaches the ground when t = 2 seconds.
The downward motion started at the vertex, when t = 1.
So the duration of the downward motion is 2 - 1 = 1 second.