Question

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 $y_0 = 0$.

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: $- 4.6\ m/s$