Question

A ball is thrown straight up with an initial velocity pf 19 m/s. Take the point of release y0=0, and the upwards to be the positive direction.
(a) Calculate the displacement at the time of 0.50 s
(b) Calculate the velocity at the time of 0.50 s
(c) Calculate the displacement at the time of 1.0 s
(d) Calculate the velocity at the time of 1.0 s
(e) Calculate the displacement at the time of 1.5 s
(f) Calculate the velocity at the time of 1.5 s
(g) Calculate the displacement at the time of 2.0 s
(h) Calculate the velocity at the time of 2.0 s

Answer 1

Answer on Question 57616, Physics, Other

Question:

A ball is thrown straight up with an initial velocity of $19\,m/s$. Take the point of release $y_0 = 0$, and the upwards to be the positive direction.

a) Calculate the displacement at the time of $0.50\,s$.

b) Calculate the velocity at the time of $0.50\,s$.

c) Calculate the displacement at the time of $1.0\,s$.

d) Calculate the velocity at the time of $1.0\,s$.

e) Calculate the displacement at the time of $1.5\,s$.

f) Calculate the velocity at the time of $1.5\,s$.

g) Calculate the displacement at the time of $2.0\,s$.

h) Calculate the velocity at the time of $2.0\,s$.

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, $y_0 = 0\,m$ is the point of release, $v_0$ is the initial velocity, $t$ is the time,

$a = g = -9.8\,\frac{m}{s^2}$ is the acceleration due to gravity.

Then, the displacement at the time of $0.50\,s$ will be:

$$y_1 = y_0 + v_0 t_1 + \frac{1}{2} a t_1^2 = 0\,m + 19\,\frac{m}{s} \cdot 0.50\,s + \frac{1}{2} \cdot \left(-9.8\,\frac{m}{s^2}\right) \cdot (0.50\,s)^2 = 8.28\,m.$$

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

$$v = v_0 + a t,$$

here, $v_{0}$ is the initial velocity, $t$ is the time, $a = g = -9.8\frac{m}{s^2}$ is the acceleration due to gravity. Then, the velocity at the time of 0.50 s will be:

$$v_{1} = v_{0} + a t_{1} = 19 \frac{m}{s} + \left(-9.8 \frac{m}{s^{2}}\right) \cdot 0.50 \, s = 14.1 \frac{m}{s}.$$

c) The displacement at the time of $1.0 \, s$ will be:

$$\begin{array}{l} y_{2} = y_{0} + v_{0} t_{2} + \frac{1}{2} a t_{2}^{2} = 0 \, m + 19 \frac{m}{s} \cdot 1.0 \, s + \frac{1}{2} \cdot \left(-9.8 \frac{m}{s^{2}}\right) \cdot (1.0 \, s)^{2} = \\ = 14.1 \, m. \end{array}$$

d) The velocity at the time of $1.0 \, s$ will be:

$$v_{2} = v_{0} + a t_{2} = 19 \frac{m}{s} + \left(-9.8 \frac{m}{s^{2}}\right) \cdot 1.0 \, s = 9.2 \frac{m}{s}.$$

e) The displacement at the time of $1.5 \, s$ will be:

$$y_{3} = y_{0} + v_{0} t_{3} + \frac{1}{2} a t_{3}^{2} = 0 \, m + 19 \frac{m}{s} \cdot 1.5 \, s + \frac{1}{2} \cdot \left(-9.8 \frac{m}{s^{2}}\right) \cdot (1.5 \, s)^{2} = 17.5 \, m.$$

f) The velocity at the time of $1.5 \, s$ will be:

$$v_{3} = v_{0} + a t_{3} = 19 \frac{m}{s} + \left(-9.8 \frac{m}{s^{2}}\right) \cdot 1.5 \, s = 4.3 \frac{m}{s}.$$

g) The displacement at the time of $2.0 \, s$ will be:

$$y_{4} = y_{0} + v_{0} t_{4} + \frac{1}{2} a t_{4}^{2} = 0 \, m + 19 \frac{m}{s} \cdot 2.0 \, s + \frac{1}{2} \cdot \left(-9.8 \frac{m}{s^{2}}\right) \cdot (2.0 \, s)^{2} = 18.4 \, m.$$

h) The velocity at the time of $2.0 \, s$ will be:

$$v_{4} = v_{0} + a t_{4} = 19 \frac{m}{s} + \left(-9.8 \frac{m}{s^{2}}\right) \cdot 2.0 \, s = -0.6 \frac{m}{s}.$$

The sign minus indicates that the ball is begin to fall.

**Answer:**

a) $y_{1} = 8.28 \, m$,

b) $y_{2} = 14.1 \, m$,

c) $y_{3} = 17.5 \, m$,

d) $y_{4} = 18.4 \, m$,

e) $v_{1} = 14.1 \, m/s$,

f) $v_{2} = 9.2 \, m/s$,

g) $v_{3} = 4.3 \, m/s$,

https://www.AssignmentExpert.com