Question

part (a) how long does it take the ball to reach its maximum point (t_{top}=0.8160 s)? correct! part (g) what is the total displacement, in meters, of the ball for its entire motion: traveling from the ground to the top and then returning to the ground? (delta y_{total}=0.000 m) correct! part (h) enter an expression for the total time of flight of the ball: the time from when it is launched to when it lands back on the ground. express this time in terms of (v_i) and (a). correct! (t_{total}=- 2v_i/a) part (i) calculate the value of (t_{total}) in seconds. (t_{total}=1.632 s) correct! part (j) enter an expression for (t_{total}) in terms of the variable (t_{top}), the time of ascent from the ground to the top of the trajectory. (t_{total}=2t_{top}) correct! part (k) express the velocity of the ball in the instant of time before it hits the ground, (v_f), in terms of (v_i), (t_{total}), and (a).

Explanation:

Step1: Recall kinematic - equation

The kinematic equation $v = v_i+at$. At the maximum - height, the velocity is $0$. The time to reach the maximum - height is $t_{top}$. We know that $0 = v_i - at_{top}$ (taking the upward direction as positive and acceleration $a$ is the acceleration due to gravity acting downwards), so $v_i=at_{top}$.

Step2: Analyze the motion for part (g)

The total displacement $\Delta y_{total}$ of an object in free - fall (going up and coming back down) can be found using the kinematic equation $\Delta y=v_it+\frac{1}{2}at^{2}$. When the object returns to the starting point, $\Delta y_{total} = 0$. The total time of flight is $t_{total}$. So, $0 = v_it_{total}-\frac{1}{2}at_{total}^{2}$. Factoring out $t_{total}$, we get $t_{total}(v_i-\frac{1}{2}at_{total}) = 0$. One solution is $t_{total}=0$ (corresponds to the initial time). The non - zero solution is $t_{total}=\frac{2v_i}{a}$.

Step3: Analyze the motion for part (h)

We already found that $t_{total}=\frac{2v_i}{a}$.

Step4: Analyze the motion for part (i)

We know $t_{total} = 2t_{top}$, so $t_{top}=\frac{t_{total}}{2}$.

Step5: Analyze the motion for part (k)

The velocity $v_f$ at a time $t$ before hitting the ground can be found using $v_f=v_i - at$. We know $t = t_{total}-\Delta t$ (where $\Delta t$ is the small time interval before hitting the ground). Since $t_{total}=\frac{2v_i}{a}$, we can also use the fact that the motion is symmetric. The velocity at a time $t$ before hitting the ground is the same as the velocity at a time $t$ after being launched in magnitude but with opposite direction in the non - dissipative case. We can use the kinematic equation $v_f=v_i - a(t_{total}-\Delta t)$. In terms of the given variables, we know that the velocity at time $t$ before hitting the ground: First, since $t_{total}=\frac{2v_i}{a}$, we have $v_f=v_i - a(\frac{2v_i}{a}-\Delta t)=v_i - 2v_i+a\Delta t=-v_i + a\Delta t$. In terms of $v_i$, $t_{total}$, and $a$, if we consider the time $t$ before hitting the ground, and we know $t_{total}=\frac{2v_i}{a}$, we can write $v_f = v_i-a(t_{total}-\Delta t)$. If we want to express it in terms of $v_i$, $t_{total}$, and $a$ without $\Delta t$ (assuming we are interested in the general case just before hitting the ground), we know that the velocity just before hitting the ground is $v_f=-v_i$ (using the symmetry of free - fall motion, the velocity when it comes back to the starting point has the same magnitude as the initial velocity but opposite direction). Mathematically, using $v = v_i+at$ with $t = t_{total}=\frac{2v_i}{a}$, $v_f=v_i+a\times\frac{2v_i}{a}=v_i - 2v_i=-v_i$.

Answer:

For part (g), the total displacement $\Delta y_{total} = 0$.
For part (h), the expression for the total time of flight is $t_{total}=\frac{2v_i}{a}$.
For part (i), $t_{top}=\frac{t_{total}}{2}$.
For part (k), $v_f=-v_i$ (velocity just before hitting the ground in terms of $v_i$, $t_{total}$, and $a$).