Question

part (d) what is the maximum height the ball reaches in meters? (h_{max}=3.265 m) correct! part (e) enter an expression for the elapsed time it takes for the ball to travel from the ground to a given height (h), and the acceleration (a). (delta t = v_i-v_f/a) incorrect! part (f) how long does it take the ball to reach its maximum point (t_{op}) in seconds? (t_{op}=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). (t_{total}=-2v_i/a) correct! part (i) calculate the value of (t_{total}) in seconds. (t_{total}=) blank grade summary deductions potential 0% 100% submissions attempts remaining: 3 4% deduction per attempt detailed view

Explanation:

Step1: Recall kinematic equations

For vertical - motion under constant acceleration (free - fall, $a=-g=- 9.8\ m/s^{2}$), the main equations are $v = v_i+at$, $y - y_i=v_it+\frac{1}{2}at^{2}$, and $v^{2}=v_{i}^{2}+2a(y - y_i)$.

Step2: Analyze part (d)

At the maximum height, the final vertical velocity $v = 0$. Using the equation $v^{2}=v_{i}^{2}+2a(y - y_i)$, with $y - y_i=h_{max}$, $v = 0$ and $a=-g$. Rearranging for $h_{max}$ gives $h_{max}=\frac{v_{i}^{2}}{2g}$. But since we are given $h_{max}=3.265\ m$, we assume this value is already calculated correctly from the initial conditions.

Step3: Analyze part (e)

We use the equation $v = v_i+at$. To find the time $t$ to reach a height $h$ with initial velocity $v_i$ and acceleration $a$, we can rewrite it as $t=\frac{v - v_i}{a}$. Here, $t=\Delta t=\frac{v - v_i}{a}$.

Step4: Analyze part (f)

At the maximum - height, $v = 0$. Using $v = v_i+at$, and solving for $t$ (time to reach the maximum - height $t_{top}$), we get $t_{top}=\frac{0 - v_i}{a}=\frac{v_i}{g}$ (since $a=-g$). Given $t_{top}=0.8160\ s$.

Step5: Analyze part (g)

The total displacement $\Delta y_{total}$ for an object that goes up and comes back down to the same level is $0$ (because the starting and ending positions are the same in the vertical direction).

Step6: Analyze part (h)

The time of flight $t_{total}$ for an object in free - fall with initial vertical velocity $v_i$ and acceleration $a=-g$ is given by the fact that the motion is symmetric. The time to go up is the same as the time to come down. Using $v = v_i+at$, when the object comes back to the starting position, the displacement $y - y_i = 0$. Using $y - y_i=v_it+\frac{1}{2}at^{2}$ with $y - y_i = 0$ and $a=-g$, we get $0 = v_it-\frac{1}{2}gt^{2}=t(v_i-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the start time) and the other is $t=\frac{2v_i}{g}$. So $t_{total}=\frac{2v_i}{a}$ (where $a=-g$).

Step7: Analyze part (i)

We need to find the value of $t_{total}$ in seconds. Since $t_{total}=\frac{2v_i}{a}$ and we know from part (f) that $t_{top}=\frac{v_i}{a}=0.8160\ s$, then $t_{total}=2t_{top}$. So $t_{total}=2\times0.8160\ s = 1.632\ s$.

Answer:

$1.632$