Question

7/8 points
calculate the displacement (in m) and velocity (in m/s) at the following times for a ball thrown straight up with an initial velocity of 15.7 m/s. take the point of release to be $y_0 = 0$. (assume the +y-axis is up. indicate the direction with the signs of your answers.)
(a) 0.595 s
displacement $square$ m
velocity $square$ m/s
(b) 1.19 s
displacement $square$ m
velocity $square$ m/s
(c) 1.79 s
displacement $square$ m
velocity $square$ m/s
(d) 2.38 s
displacement $square$ m
velocity $square$ m/s
details
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osuniphys1 3.5.p.066.

Explanation:

Step1: List given values

Initial velocity $v_0 = 15.7\ \text{m/s}$, initial position $y_0 = 0$, acceleration $a = -g = -9.81\ \text{m/s}^2$

Step2: Displacement formula

Displacement: $y(t) = v_0 t + \frac{1}{2} a t^2$

Step3: Velocity formula

Velocity: $v(t) = v_0 + a t$

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Part (a): $t=0.595\ \text{s}$

Step1: Calculate displacement

Substitute $t=0.595$ into displacement formula:
$y(0.595) = 15.7(0.595) + \frac{1}{2}(-9.81)(0.595)^2$
$y(0.595) = 9.3415 - 1.743 = 7.60\ \text{m}$

Step2: Calculate velocity

Substitute $t=0.595$ into velocity formula:
$v(0.595) = 15.7 + (-9.81)(0.595)$
$v(0.595) = 15.7 - 5.837 = 9.86\ \text{m/s}$

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Part (b): $t=1.19\ \text{s}$

Step1: Calculate displacement

Substitute $t=1.19$ into displacement formula:
$y(1.19) = 15.7(1.19) + \frac{1}{2}(-9.81)(1.19)^2$
$y(1.19) = 18.683 - 6.964 = 11.72\ \text{m}$

Step2: Calculate velocity

Substitute $t=1.19$ into velocity formula:
$v(1.19) = 15.7 + (-9.81)(1.19)$
$v(1.19) = 15.7 - 11.674 = 4.03\ \text{m/s}$

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Part (c): $t=1.79\ \text{s}$

Step1: Calculate displacement

Substitute $t=1.79$ into displacement formula:
$y(1.79) = 15.7(1.79) + \frac{1}{2}(-9.81)(1.79)^2$
$y(1.79) = 28.103 - 15.704 = 12.40\ \text{m}$

Step2: Calculate velocity

Substitute $t=1.79$ into velocity formula:
$v(1.79) = 15.7 + (-9.81)(1.79)$
$v(1.79) = 15.7 - 17.560 = -1.86\ \text{m/s}$

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Part (d): $t=2.38\ \text{s}$

Step1: Calculate displacement

Substitute $t=2.38$ into displacement formula:
$y(2.38) = 15.7(2.38) + \frac{1}{2}(-9.81)(2.38)^2$
$y(2.38) = 37.366 - 27.882 = 9.48\ \text{m}$

Step2: Calculate velocity

Substitute $t=2.38$ into velocity formula:
$v(2.38) = 15.7 + (-9.81)(2.38)$
$v(2.38) = 15.7 - 23.348 = -7.65\ \text{m/s}$

Answer:

(a) displacement: $7.60\ \text{m}$, velocity: $9.86\ \text{m/s}$
(b) displacement: $11.72\ \text{m}$, velocity: $4.03\ \text{m/s}$
(c) displacement: $12.40\ \text{m}$, velocity: $-1.86\ \text{m/s}$
(d) displacement: $9.48\ \text{m}$, velocity: $-7.65\ \text{m/s}$