Question

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:

a) Maximum height will be $h=300+\Delta y=305.10m$.

b) $y(4s)=261.6m$.

c) The time before it hits the ground is $t = 6.87s.$.

Explanation:

Initial position: $y_{i}=300m$

Initial velocity: $v_{i}=10.0 \frac{m}{s}$

Gravity's acceleration: $g=-9.8 \frac{m}{s^{2} }$

a) From Kinematics: $v^2=v_0^2+2g\Delta y$

⇒ $\Delta y=\frac{v^2-v_0^2}{2g}$

When the coin reaches it's maximum height, it's velocity will be $v=0 \frac{m}{s}$

⇒ $\Delta y=\frac{v_0^2}{2g}$ ⇒ $\Delta y=5.10m$.

Maximum height will be $h=300+\Delta y=305.10m$.

b) From Kinematics, the movement law will be:

$y(t)=y_0 + v_0t + \frac{1}{2}gt^2$

where $y_0=300m$ is the position where the coin is dropped.

At $t=4s$:

$y(4s)=261.6m$.

c) Again, we have that:

$y(t)=y_0 + v_0t + \frac{1}{2}gt^2$

Now, if $y$ final is $y(t_f)=0$, then

$y_0 + v_0t_f + \frac{1}{2}gt_f^2=0$.

We need to use the resolvent to solve this equation:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-v_0\pm \sqrt{v_0^2-2g y_0}}{g}$

⇒ $t = 6.87s.$.