Question

13) A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Answer 1

Answer on Question #58682 – Physics/Mechanics – Relativity

A coin is dropped from a hot-air balloon that is $300\,\mathrm{m}$ above the ground and rising at $10.0\,\mathrm{m/s}$ upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Solution:

General formula for position is:

$$x = x_0 + v_0 \cdot t + \frac{a \cdot t^2}{2}$$

Where $x_0$ is starting distance, $v_0$ is starting velocity and $a$ is acceleration. For this task, $x_0 = 300\,\mathrm{m}$, $v_0 = 10\,\frac{\mathrm{m}}{\mathrm{s}}$ and $a = -g = -9.8\,\frac{\mathrm{m}}{\mathrm{s}^2}$, because acceleration is directed downwards.

a) To find maximum height reached, one can derive $x(t)$ by $t$ and get:

$$\frac{d}{dt} x(t) = v_0 + a t$$

Extremum of function demands first derivative to be equal to zero, thus:

$$v_0 + a t = 0; \, t = -\frac{v_0}{a} = -\frac{10\,\frac{\mathrm{m}}{\mathrm{s}}}{9.8\,\frac{\mathrm{m}}{\mathrm{s}^2}} \approx 1.02\,\mathrm{s}$$

Maximum height will be reached at $t = 1.02\,\mathrm{s}$ and will be equal to:

$$x(t = 1.02\,\mathrm{s}) = 300 + 10 \cdot 1.02 - \frac{9.8 \cdot 1.02^2}{2} = 305.102 = 305.1(\mathrm{m})$$

b) Velocity after 4.00 s after release can be calculated as follows:

$$v(t = 4.00\,\mathrm{s}) = v_0 + a t = 10 - 9.8 \cdot 4 = -29.2 \left(\frac{\mathrm{m}}{\mathrm{s}}\right)$$

negative sign means coin is travels downwards. Position after 4.00 s after release:

$$x(t = 4.00\,\mathrm{s}) = x_0 + v_0 t + \frac{a \cdot t^2}{2} = 300 + 10 \cdot 4.00 - \frac{9.8 \cdot 4^2}{2} = 261.6(\mathrm{m})$$

above the ground.

c) Coin hitting the ground means it's position is $0\,\mathrm{m}$. To get time, one can solve the equation:

$$0 = x_0 + v_0 t + \frac{a \cdot t^2}{2}$$ $$D = b^2 - 4ac = v_0^2 - 4 \cdot \frac{a \cdot x_0}{2} = v_0^2 - 2ax_0 = 100 - 2 \cdot (-9.8) \cdot 300 = 5980$$ $$t = \frac{-v_0 \pm \sqrt{D}}{a} = \frac{-10 \pm 77.33}{9.8}$$

There are two roots, one of which is negative – but because time can't be negative, that root is not physical one, so it's neglected. The one that's left:

$$t = 6.87\,\mathrm{s}$$

Answer:

a) Maximum height reached $x = 305.1(\mathrm{m})$.

b) Velocity after $4.00 \, \text{s} \, v = -29.2 \left( \frac{m}{s} \right)$, position after $4.00 \, \text{s} \, x = 261.6 \, (m)$

c) Total time of flight $t = 6.87 \, \text{s}$.

https://www.AssignmentExpert.com