[Q1] What is the time it takes for the object to ascend to its highest point?

The motion is analyzed by considering the ascent and descent phases separately, both of which occur under constant but different accelerations. Let $m$ be the object's mass and $g$ be the acceleration due to gravity. The air resistance force is constant, $F_R = \frac{1}{5}mg$.

1. Determine Accelerations

• Ascent: Gravity and air resistance both act downwards, opposing the initial upward velocity. The net force is $F_{net, up} = mg + F_R$. $$ma_{up} = mg + \frac{1}{5}mg = \frac{6}{5}mg$$ The magnitude of the upward acceleration (deceleration) is $a_{up} = \frac{6}{5}g$.
• Descent: Gravity acts downwards, while air resistance acts upwards, opposing the downward velocity. The net force is $F_{net, down} = mg - F_R$. $$ma_{down} = mg - \frac{1}{5}mg = \frac{4}{5}mg$$ The magnitude of the downward acceleration is $a_{down} = \frac{4}{5}g$.

2. Relate Height and Time Let $H$ be the maximum height, $t_{up}$ the ascent time, and $t_{down}$ the descent time. Using the kinematic equation for displacement under constant acceleration, we relate the height to the times for each phase.

• For ascent, starting with initial velocity $v_0$ and ending with $v_f=0$: $H = \frac{v_0+v_f}{2}t_{up} = \frac{v_0}{2}t_{up}$. Since $v_0 = a_{up}t_{up}$, we have $H = \frac{1}{2}a_{up}t_{up}^2$.
• For descent, starting from rest ($v_i=0$): $H = \frac{1}{2}a_{down}t_{down}^2$.

Equating the two expressions for the maximum height $H$:

$$\frac{1}{2}a_{up}t_{up}^2 = \frac{1}{2}a_{down}t_{down}^2$$ $$t_{down} = t_{up} \sqrt{\frac{a_{up}}{a_{down}}}$$

3. Solve for Ascent Time The total time of flight is given as $T = t_{up} + t_{down} = 6$ s. Substituting the expression for $t_{down}$:

$$T = t_{up} + t_{up} \sqrt{\frac{a_{up}}{a_{down}}} = t_{up} \left(1 + \sqrt{\frac{a_{up}}{a_{down}}}\right)$$

Solving for the ascent time, $t_{up}$:

$$t_{up} = \frac{T}{1 + \sqrt{\frac{a_{up}}{a_{down}}}}$$

4. Substitute Values First, find the ratio of the accelerations:

$$\frac{a_{up}}{a_{down}} = \frac{\frac{6}{5}g}{\frac{4}{5}g} = \frac{6}{4} = \frac{3}{2}$$

Now, substitute the values of $T$ and the acceleration ratio into the expression for $t_{up}$:

$$t_{up} = \frac{6 \text{ s}}{1 + \sqrt{\frac{3}{2}}} = \frac{6}{1 + \frac{\sqrt{6}}{2}} = \frac{12}{2 + \sqrt{6}}$$

To simplify, we can rationalize the denominator:

$$t_{up} = \frac{12(2 - \sqrt{6})}{(2 + \sqrt{6})(2 - \sqrt{6})} = \frac{12(2 - \sqrt{6})}{4 - 6} = 6(\sqrt{6} - 2) \text{ s}$$

Calculating the numerical value:

$$t_{up} \approx 2.697 \text{ s}$$