assume air resistance is negligible unless otherwise stated. 41. calculate the displacement and velocity at…

assume air resistance is negligible unless otherwise stated. 41. calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. take the point of release to be $y_0 = 0$.
Answer
Explanation:
Step1: Identify the kinematic equations
The equations for vertical - motion under gravity are $y = y_0+v_0t-\frac{1}{2}gt^2$ and $v = v_0 - gt$, where $y_0$ is the initial position, $v_0$ is the initial velocity, $t$ is the time, $g = 9.8\ m/s^2$ is the acceleration due to gravity, $y$ is the displacement, and $v$ is the velocity. Here, $y_0 = 0$ and $v_0=15.0\ m/s$.
Step2: Calculate for $t = 0.500\ s$
For displacement: $y_1=0 + 15\times0.5-\frac{1}{2}\times9.8\times(0.5)^2$ $y_1=7.5 - 1.225=6.275\ m$ For velocity: $v_1=15-9.8\times0.5$ $v_1=15 - 4.9 = 10.1\ m/s$
Step3: Calculate for $t = 1.00\ s$
For displacement: $y_2=0 + 15\times1-\frac{1}{2}\times9.8\times(1)^2$ $y_2=15 - 4.9 = 10.1\ m$ For velocity: $v_2=15-9.8\times1$ $v_2=15 - 9.8 = 5.2\ m/s$
Step4: Calculate for $t = 1.50\ s$
For displacement: $y_3=0 + 15\times1.5-\frac{1}{2}\times9.8\times(1.5)^2$ $y_3=22.5-11.025 = 11.475\ m$ For velocity: $v_3=15-9.8\times1.5$ $v_3=15 - 14.7 = 0.3\ m/s$
Step5: Calculate for $t = 2.00\ s$
For displacement: $y_4=0 + 15\times2-\frac{1}{2}\times9.8\times(2)^2$ $y_4=30 - 19.6 = 10.4\ m$ For velocity: $v_4=15-9.8\times2$ $v_4=15 - 19.6=-4.6\ m/s$
Answer:
(a) Displacement: $6.275\ m$, Velocity: $10.1\ m/s$ (b) Displacement: $10.1\ m$, Velocity: $5.2\ m/s$ (c) Displacement: $11.475\ m$, Velocity: $0.3\ m/s$ (d) Displacement: $10.4\ m$, Velocity: $- 4.6\ m/s$