Quadratic Mechanics

Relevant wiki: Equations Of Motion

I would solve this problem by appealing to symmetry.

Let the starting point be O, the point at height $h$ be A, the highest point is B. Hence, the ball travels in the order OABAO.

By considering the symmetry of the situation, note that time to travel OA = time to travel AO

The time required by the question = time to travel OA + time to travel OABA

= time to travel AO + time to travel OABA

= time to travel OABAO

Of course, the speed at which the ball returns to O is the same as when it leaves, i.e. 2m/s.

Hence, we just need to apply $v = u + at$ to get:

$-2 = 2 + (-10)t$

t = 0.4 second.

Using the projectile motion formula $h=-\frac{gt^2}{2}+vt+h_0$ , we get the function $h=-5t^2+2t$ . Seeing this is a quadratic function, we can notice that the sum of the 2 times the ball reaches the height h is simply the time it hits the ground. Setting this equal to 0 and factoring, we get $t(2-5t)=0 \implies t=0, \frac{2}{5}$ Therefore, the answer is $\frac{2}{5} = \boxed{0.4}$

Pretty simple.

We know, $h=ut-(gt^2)/2$

After rearranging terms, you get the quadratic,

$-gt^2+2ut-2h=0$

Applying Vieta's Formula,

The sum of the times at which it is seen is $2u/g$

Let's take a suggestion from the wording of the question and assume that the result holds for any height to which the ball reaches. This includes the maximum height, which is reached as the speed goes to zero. This happens at $t = \frac{2}{10}$ . Doubling it gives $\frac{4}{10}$ .

The symmetry of the two times $t_1$ and $t_2$ at which the ball is seen at $h$ implies that they are equally apart from the time it reaches the top of its trajectory. We can quickly solve for that time:

$\displaystyle \begin{aligned} v_f & = v_0 + at \\ 0 & = 2 - 10t \\ t & = 0.2 \ \text{seconds} \end{aligned}$

Now let $t_0$ be the length of time between $t = 0.2$ and $t_1$ and $t_2$ . We can then write $t_1$ as $0.2 - t_0$ and $t_2$ and $0.2 + t_0$ .

Hence, the sum of the times the ball is seen at $h$ is

$\displaystyle t_1 + t_2 = (0.2 - t_0) + (0.2 + t_0) = \boxed{0.4}$

The height of the ball as a function of time is given by the equation:
$y=v_{0}t-\frac{1}{2}gt^{2}\Rightarrow$
$y=2t-5t^{2}$
For $y=h$ we get:
$5t^{2}-2t+h=0$
The moments that the height of the ball is $h$ are the roots of this equations.
So their sum, according to Vieta, is:
$S=-\frac{b}{a}\Rightarrow$
$S=-\frac{-2}{5}\Rightarrow$
$\boxed{S=\frac{2}{5}}$

In very basic terms, we want to find the time of flight of vertical projectile above height 'h' with initial velocity 2 m/s. We can simply ignore the fact that the projectile has traveled 'h' height initially. We can directly find its time of flight.

Using the formula $T = \dfrac{2usin\theta}{g}$ and plugging values of u=2, $\theta = 90^\circ$ and g=10, we get

Total Time of Flight $\boxed {0.4 s}$

• Vf = Vi + a.t

• h= 0 (flight is parabolic and symmetric, so the two times added together at height h will be the same, irrespective of h)

• Vf = -2 (vector quantity, ball falling is negative)
• Vi = 2 (initial velocity)

• a = -10 (gravity is negative)

• -2 = 2 + -10.t

• -4 = -10.t
• -4/-10 = t
• 4/10 = t
• 0.4 = t

As we know v= u +gt

Where v is the final velocity and u is the initial velocity. So, now let's put in the values.

0=2 -10t. [v becomes 0 when the ball reaches maximum height]

=> t=1/5 = 0.2secs

Now, as Time of Ascent = Time of Descent

=> Total time = 0.4 secs