Question

A rock is thrown straight down with an initial velocity of 10.5 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70m above water. Take upwards to be the positive direction.

(a) Calculate the displacement at a time of 1.0 s
(b) Calculate the velocity at a time of 1.0 s
(c) Calculate the displacement at a time of 1.5 s
(d) Calculate the velocity at a time of 1.5 s
(e) Calculate the displacement at a time of 2.0 s
(f) Calculate the velocity at a time of 2.0 s
(g) Calculate the displacement at a time of 2.5 s
(h) Calculate the velocity at a time of 2.5 s

Answer 1

Answer on Question 57617, Physics, Other

Question:

A rock is thrown straight down with an initial velocity of $10.5 \, \text{m/s}$ from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is $70 \, \text{m}$ above water. Take upwards to be the positive direction.

a) Calculate the displacement at a time of $1.0 \, \text{s}$.

b) Calculate the velocity at a time of $1.0 \, \text{s}$.

c) Calculate the displacement at a time of $1.5 \, \text{s}$.

d) Calculate the velocity at a time of $1.5 \, \text{s}$.

e) Calculate the displacement at a time of $2.0 \, \text{s}$.

f) Calculate the velocity at a time of $2.0 \, \text{s}$.

g) Calculate the displacement at a time of $2.5 \, \text{s}$.

h) Calculate the velocity at a time of $2.5 \, \text{s}$.

Solution:

a) In order to find the displacement we can use the formula:

$$y = y_0 + v_0 t + \frac{1}{2} a t^2,$$

here, $y_0 = 0 \, \text{m}$ is the point of release, $v_0 = -10.5 \, \text{m/s}$ is the initial velocity, $t$ is the time, $a = g = -9.8 \frac{\text{m}}{\text{s}^2}$ is the acceleration due to gravity.

Then, the displacement at the time of $1.0 \, \text{s}$ will be:

$$y_1 = y_0 + v_0 t_1 + \frac{1}{2} a t_1^2 = 0 \, \text{m} + \left(-10.5 \, \frac{\text{m}}{\text{s}}\right) \cdot 1.0 \, \text{s} + \frac{1}{2} \cdot \left(-9.8 \, \frac{\text{m}}{\text{s}^2}\right) \cdot (1.0 \, \text{s})^2 = -15.4 \, \text{m}.$$

b) In order to find the velocity we can use the formula:

$$v = v_0 + a t,$$

here, $v_0 = -10.5 \, \text{m/s}$ is the initial velocity, $t$ is the time, $a = g = -9.8 \frac{\text{m}}{\text{s}^2}$ is the acceleration due to gravity.

Then, the velocity at the time of $1.0 \, \text{s}$ will be:

$$v _ {1} = v _ {0} + a t _ {1} = \left(- 1 0. 5 \frac {m}{s}\right) + \left(- 9. 8 \frac {m}{s ^ {2}}\right) \cdot 1. 0 s = - 2 0. 3 \frac {m}{s}.$$

c) The displacement at the time of $1.5s$ will be:

$$\begin{array}{l} y _ {2} = y _ {0} + v _ {0} t _ {2} + \frac {1}{2} a t _ {2} ^ {2} = 0 m + \left(- 1 0. 5 \frac {m}{s}\right) \cdot 1. 5 s + \frac {1}{2} \cdot \left(- 9. 8 \frac {m}{s ^ {2}}\right) \cdot (1. 5 s) ^ {2} = \\ = - 2 6. 8 m. \\ \end{array}$$

d) The velocity at the time of $1.5s$ will be:

$$v _ {2} = v _ {0} + a t _ {2} = \left(- 1 0. 5 \frac {m}{s}\right) + \left(- 9. 8 \frac {m}{s ^ {2}}\right) \cdot 1. 5 s = - 2 5. 2 \frac {m}{s}.$$

e) The displacement at the time of $2.0s$ will be:

$$\begin{array}{l} y _ {3} = y _ {0} + v _ {0} t _ {3} + \frac {1}{2} a t _ {3} ^ {2} = 0 m + \left(- 1 0. 5 \frac {m}{s}\right) \cdot 2. 0 s + \frac {1}{2} \cdot \left(- 9. 8 \frac {m}{s ^ {2}}\right) \cdot (2. 0 s) ^ {2} = \\ = - 4 0. 6 m. \\ \end{array}$$

f) The velocity at the time of $2.0s$ will be:

$$v _ {3} = v _ {0} + a t _ {3} = \left(- 1 0. 5 \frac {m}{s}\right) + \left(- 9. 8 \frac {m}{s ^ {2}}\right) \cdot 2. 0 s = - 3 0. 1 \frac {m}{s}.$$

g) The displacement at the time of $2.5s$ will be:

$$\begin{array}{l} y _ {4} = y _ {0} + v _ {0} t _ {4} + \frac {1}{2} a t _ {4} ^ {2} = 0 m + \left(- 1 0. 5 \frac {m}{s}\right) \cdot 2. 5 s + \frac {1}{2} \cdot \left(- 9. 8 \frac {m}{s ^ {2}}\right) \cdot (2. 5 s) ^ {2} = \\ = - 5 6. 9 m. \\ \end{array}$$

h) The velocity at the time of $2.5s$ will be:

$$v _ {4} = v _ {0} + a t _ {4} = \left(- 1 0. 5 \frac {m}{s}\right) + \left(- 9. 8 \frac {m}{s ^ {2}}\right) \cdot 2. 5 s = - 3 5. 0 \frac {m}{s}.$$

Answer:

a) $y_{1} = -15.4m$ , c) $y_{2} = -26.8m$ , e) $y_{3} = -40.6m$ , g) $y_{4} = -56.9m$

b) $v_{1} = -20.3 \, \text{m/s}$ , d) $v_{2} = -25.2 \, \text{m/s}$ , f) $v_{3} = -30.1 \, \text{m/s}$ , h) $v_{4} = -35.0 \, \text{m/s}$ .

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