Question

a boy is standing on a pole of height 14.7m throws a stone upwards. it moves in a vertical line slightly away from the pole and falls on the ground. its equation of motion in meters and seconds is x=9.8t-4.9t^2. find time taken for upward and downward motions. also find the maximum height reached by the stone from the ground.

Answer 1

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:

$x(t) = -4.9t^{2} + 9.8t$

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 $x(t) = at^{2} + bt + c$

The vertex is the point $(t_{v}, x(t_{v}))$, in which

$t_{v} = -\frac{b}{2a}$

In this question:

$x(t) = -4.9t^{2} + 9.8t$

So $a = -4.9, b = 9.8$

Vertex:

$t_{v} = -\frac{9.8}{2*(-4.9)} = 1$

The time taken for the upward motion is 1 second.

$x(t_{v}) = x(1) = 9.8*1 - 4.9*(1)^{2} = 4.9$

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

$-4.9t^{2} + 9.8t = 0$

$4.9t^{2} - 9.8t = 0$

$4.9t(t - 2) = 0$

$4.9t = 0$

$t = 0$

Or

$t - 2 = 0$

$t = 2$

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.