Question

1. calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the verrazano narrows bridge in new york city. the roadway of this bridge is 70.0 m above the water. 43. a basketball referee tosses the ball straight up for the starting tip - off. at what minimum velocity must a

Explanation:

Step1: Identify the kinematic equations

The equations for displacement $y - y_0=v_0t+\frac{1}{2}at^2$ and velocity $v = v_0+at$ will be used. Here, $y_0 = 0$, $v_0=14.0\ m/s$, and $a = g= 9.8\ m/s^2$.

Step2: Calculate displacement for (a) $t = 0.500\ s$

$y - y_0=v_0t+\frac{1}{2}at^2$. Substitute $v_0 = 14.0\ m/s$, $a = 9.8\ m/s^2$, and $t = 0.500\ s$.
$y=14.0\times0.500+\frac{1}{2}\times9.8\times(0.500)^2=7.00 + 1.225=8.225\ m$
$v=v_0+at=14.0+9.8\times0.500=14.0 + 4.9=18.9\ m/s$

Step3: Calculate displacement for (b) $t = 1.00\ s$

$y=14.0\times1.00+\frac{1}{2}\times9.8\times(1.00)^2=14.0+4.9 = 18.9\ m$
$v=v_0+at=14.0+9.8\times1.00=14.0 + 9.8=23.8\ m/s$

Step4: Calculate displacement for (c) $t = 1.50\ s$

$y=14.0\times1.50+\frac{1}{2}\times9.8\times(1.50)^2=21.0+11.025 = 32.025\ m$
$v=v_0+at=14.0+9.8\times1.50=14.0+14.7 = 28.7\ m/s$

Step5: Calculate displacement for (d) $t = 2.00\ s$

$y=14.0\times2.00+\frac{1}{2}\times9.8\times(2.00)^2=28.0 + 19.6=47.6\ m$
$v=v_0+at=14.0+9.8\times2.00=14.0+19.6 = 33.6\ m/s$

Step6: Calculate displacement for (e) $t = 2.50\ s$

$y=14.0\times2.50+\frac{1}{2}\times9.8\times(2.50)^2=35.0+30.625 = 65.625\ m$
$v=v_0+at=14.0+9.8\times2.50=14.0+24.5 = 38.5\ m/s$

Answer:

(a) Displacement: $8.225\ m$, Velocity: $18.9\ m/s$
(b) Displacement: $18.9\ m$, Velocity: $23.8\ m/s$
(c) Displacement: $32.025\ m$, Velocity: $28.7\ m/s$
(d) Displacement: $47.6\ m$, Velocity: $33.6\ m/s$
(e) Displacement: $65.625\ m$, Velocity: $38.5\ m/s$