Question

1. 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.

Explanation:

Step1: Analyze the initial - velocity and position

The initial velocity of the coin $v_0 = 10.0\ m/s$ (upward) and the initial height $y_0=300\ m$. At the maximum - height, the final velocity $v = 0$. We use the kinematic equation $v^{2}-v_{0}^{2}=-2g\Delta y$.

Step2: Calculate the additional height $\Delta y$

From $v^{2}-v_{0}^{2}=-2g\Delta y$, we can solve for $\Delta y$. Given $v = 0$, $v_0 = 10.0\ m/s$ and $g = 9.8\ m/s^{2}$, we have $\Delta y=\frac{v_{0}^{2}}{2g}=\frac{10^{2}}{2\times9.8}=\frac{100}{19.6}\approx5.1\ m$.
The maximum height $y_{max}=y_0+\Delta y=300 + 5.1=305.1\ m$.

Step3: Use kinematic equations for part (b)

The kinematic equations for position $y - y_0=v_0t-\frac{1}{2}gt^{2}$ and velocity $v = v_0 - gt$.
For $t = 4.00\ s$, $y=y_0+v_0t-\frac{1}{2}gt^{2}=300+10\times4-\frac{1}{2}\times9.8\times4^{2}=300 + 40-78.4=261.6\ m$.
$v = v_0 - gt=10-9.8\times4=10 - 39.2=-29.2\ m/s$ (the negative sign indicates the velocity is downward).

Step4: Use kinematic equation for part (c)

When the coin hits the ground, $y = 0$. Using the equation $y - y_0=v_0t-\frac{1}{2}gt^{2}$, we get $0 - 300=10t-4.9t^{2}$, or $4.9t^{2}-10t - 300 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 4.9$, $b=-10$, and $c=-300$.
$t=\frac{10\pm\sqrt{(-10)^{2}-4\times4.9\times(-300)}}{2\times4.9}=\frac{10\pm\sqrt{100 + 5880}}{9.8}=\frac{10\pm\sqrt{5980}}{9.8}=\frac{10\pm77.3}{9.8}$.
We take the positive root $t=\frac{10 + 77.3}{9.8}=\frac{87.3}{9.8}\approx8.91\ s$.

Answer:

(a) $305.1\ m$
(b) Position: $261.6\ m$, Velocity: $- 29.2\ m/s$
(c) $8.91\ s$