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part (a) what is the velocity of the ball, in meters per second, when i…

Question

part (a) what is the velocity of the ball, in meters per second, when it reaches the top, or highest point, of its trajectory? (v_{top}=0.000 m/s) correct! part (c) enter an expression for the height of the ball as a function of time in terms of the initial velocity (v_i) and the acceleration (a) and the elapsed time (t). (h(t)=v_it + 1/2at^{2}) correct! 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) that height (v_h), and the initial velocity (v_i), the velocity the ball has at (delta t = v_i - v_h/a) incorrect! part (f) how long does it take the ball to reach its maximum point (t_{top}) in seconds? (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

Explanation:

Step1: Analyze motion - equation for height

The height of an object in vertical - motion under constant acceleration is given by the kinematic equation $h(t)=v_it+\frac{1}{2}at^{2}$, where $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

Step2: Find time to reach maximum height

At the maximum - height, the final velocity $v_f = 0$. Using the kinematic equation $v_f=v_i + at$, we can solve for the time $t$ to reach the maximum height. Rearranging gives $t=\frac{v_f - v_i}{a}=\frac{0 - v_i}{a}=-\frac{v_i}{a}$. The time to reach the maximum height from the ground is $t_{up}=-\frac{v_i}{a}$.

Step3: Find total time of flight

The motion is symmetric. The total time of flight $t_{total}$ from launch to landing is twice the time it takes to reach the maximum height. So $t_{total}=-\frac{2v_i}{a}$.

Answer:

$t_{total}=-\frac{2v_i}{a}$