Initial velocity of a body when the distance travelled by it in the last second before reaching its maximum height is $5\ \textrm{m}$ [closed]

Question

I stumbled upon this problem recently, and it has been haunting me for more than a day at this point. The question is as follows:-

When a ball is thrown up, it reaches a maximum height of $h$, travelling a distance of $5\ \textrm{m}$ in the second before reaching $h$. Find the magnitude of velocity $u$, with which it should be thrown up to attain such a physical configuration. (Ignore air resistance)

If possible, I would like to know both the kinematical and the energy-work approach to this question.

Note: I am currently in Grade 9, therefore a level-appropriate answer would be much appreciated.

Answer 1

The question has either an infinite number of answers, or no answer.

It's convenient to follow the motion for the one second AFTER the ball reaches the peak of its trajectory. (Or run time backwards for the last second BEFORE the peak) Symmetry tells us the two views are identical.

The ball starts at zero vertical velocity, and drops $5$ metres in $1$ second.

If $g$ is taken as $10\text{ m/s/s}$ as a convenient approximation, then the $5$ metre drop is true for all drops, no matter how far.

If $g$ is taken as a more accurate number, like $9.8$, or $9.801$, then the specified drop of $5$ metres in$1$ second is impossible.

Answer 2

Assume that the velocity of the body right before the last second starts is V. It decreases to zero in 1s with an acceleration of g, about $10m/s^2$. So, V must be 10 m/s. Alo, the time to drop from that maximum height until it reaches 10m/s is also 1 second. And the distance traveled during that second, as for any first second of free fall, is about 5m. So your conditions seem to be satisfied for any vertical motion in gravitational field with g about $10m/s^2$.
Possibly you did not write the question exactly as it is given.

Energy conservation (or work-energy theorem) cannot help in these problems when time is involved.

Answer 3

Supposing the ball is thrown up vertically (otherwise the answer is undefined because the problem has not been given enought parameters): Acceleration along $z$ is $a=-g=\ddot z$ where $g\approx 9.806$ m/s$^2$. Velocity is the first integral, $\dot z = \int a dt=-gt+u$. Altitude is the second integral $$z(t)=\int \dot z dt = -gt^2/2+ut+z_0$$ Assuming altitude is measured relative to $z_0=0$, so the maximum height is where this parabola has its maximum and where $\dot z=0$, i.e., where $-gt+u=0$, so $t(h)=u/g$, so $h=-g(u/g)^2/2+uu/g=u^2/(2g)$ is the maximum altitude. $\Delta t=1$ s before $t(h)$ was at $t=u/g-\Delta t$, and the altitude was $\Delta z=5$m lower, plugging both into $z(t)$ gives $$h-\Delta h = -g(t(h)-\Delta t)^2/2+u(t(h)-\Delta t);$$ $$u^2/(2g)-\Delta h = -g(u/g-\Delta t)^2/2+u(u/g-\Delta t);$$ On both sides of the equation the terms $u^2/(2g)$ cancel and $\Delta h=g (\Delta t)^2/2$ remains, which is invalid because $g\neq 10 m/s^2$, so the requirement cannot be met with a vertical throw.

Answer 4

We know

$$S=ut+\frac{1}{2}at^2$$

Substitute $S=h$, $t=t_{f}$

$$h=ut_{f}+\frac{1}{2}at_{f}^2$$

Use the quadratic formula, to solve for t setting the discriminant to be zero, since because this is the max height, there is only 1 time at which the projectile reaches h. (This can also be done setting v=0 into the velocity equation, finding the time at which the velocity is zero and then substituting it into our expression for $t_{f}$).

Doing so gives $$t_{f} = -\frac{u}{a}$$

Substitute back into our original equation:

$$h=u[-\frac{u}{a}]+\frac{1}{2}a[-\frac{u}{a}]^2$$

So this equation means, at a time of -$\frac{u}{a}$ the projectile reaches a height h

We also know that 1 second earlier, the projectile travels 5m to get to h, so.. $(s,t) = (h-5,t_{f}-1)$

$$(h-5)=u[-\frac{u}{a}-1]+\frac{1}{2}a[-\frac{u}{a}-1]^2$$

We now have the following simultaneous equations, with 2 unknowns, h and u.

$$h=u[-\frac{u}{a}]+\frac{1}{2}a[-\frac{u}{a}]^2$$

$$(h-5)=u[-\frac{u}{a}-1]+\frac{1}{2}a[-\frac{u}{a}-1]^2$$

I'll leave it upto you to solve them. However you should be able to easily solve for u