Calculate The Displacement And Velocity At The Following Times For A Ball Thrown Straight Up With An Initial Velocity Of 15.0 M/s. Assume The Point Of Release Is $y_0 = 0$.(a) 0.500 Seconds (b) 1.00 Seconds (c) 1.50 Seconds (d) 2.00 Seconds

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Introduction

When a ball is thrown straight up into the air, it follows a parabolic trajectory under the influence of gravity. The motion of the ball can be described using the equations of motion, which relate the position, velocity, and acceleration of the ball at any given time. In this article, we will calculate the displacement and velocity of a ball thrown straight up with an initial velocity of 15.0 m/s at different times.

Equations of Motion

The equations of motion for an object under constant acceleration are given by:

  • Position: y=y0+v0t+12at2y = y_0 + v_0t + \frac{1}{2}at^2
  • Velocity: v=v0+atv = v_0 + at
  • Acceleration: a=−ga = -g (where gg is the acceleration due to gravity, which is 9.8 m/s^2)

Calculating Displacement and Velocity

We will use the equations of motion to calculate the displacement and velocity of the ball at different times.

(a) 0.500 seconds

At t=0.500t = 0.500 seconds, we can calculate the displacement and velocity of the ball using the equations of motion.

  • Displacement: y=0+(15.0 m/s)(0.500 s)+12(−9.8 m/s2)(0.500 s)2y = 0 + (15.0\, \text{m/s})(0.500\, \text{s}) + \frac{1}{2}(-9.8\, \text{m/s}^2)(0.500\, \text{s})^2 y=7.5 m−1.225 m=6.275 my = 7.5\, \text{m} - 1.225\, \text{m} = 6.275\, \text{m}
  • Velocity: v=15.0 m/s+(−9.8 m/s2)(0.500 s)v = 15.0\, \text{m/s} + (-9.8\, \text{m/s}^2)(0.500\, \text{s}) v=15.0 m/s−4.9 m/s=10.1 m/sv = 15.0\, \text{m/s} - 4.9\, \text{m/s} = 10.1\, \text{m/s}

(b) 1.00 seconds

At t=1.00t = 1.00 seconds, we can calculate the displacement and velocity of the ball using the equations of motion.

  • Displacement: y=0+(15.0 m/s)(1.00 s)+12(−9.8 m/s2)(1.00 s)2y = 0 + (15.0\, \text{m/s})(1.00\, \text{s}) + \frac{1}{2}(-9.8\, \text{m/s}^2)(1.00\, \text{s})^2 y=15.0 m−4.9 m=10.1 my = 15.0\, \text{m} - 4.9\, \text{m} = 10.1\, \text{m}
  • Velocity: v=15.0 m/s+(−9.8 m/s2)(1.00 s)v = 15.0\, \text{m/s} + (-9.8\, \text{m/s}^2)(1.00\, \text{s}) v=15.0 m/s−9.8 m/s=5.2 m/sv = 15.0\, \text{m/s} - 9.8\, \text{m/s} = 5.2\, \text{m/s}

(c) 1.50 seconds

At t=1.50t = 1.50 seconds, we can calculate the displacement and velocity of the ball using the equations of motion.

  • Displacement: y=0+(15.0 m/s)(1.50 s)+12(−9.8 m/s2)(1.50 s)2y = 0 + (15.0\, \text{m/s})(1.50\, \text{s}) + \frac{1}{2}(-9.8\, \text{m/s}^2)(1.50\, \text{s})^2 y=22.5 m−11.025 m=11.475 my = 22.5\, \text{m} - 11.025\, \text{m} = 11.475\, \text{m}
  • Velocity: v=15.0 m/s+(−9.8 m/s2)(1.50 s)v = 15.0\, \text{m/s} + (-9.8\, \text{m/s}^2)(1.50\, \text{s}) v=15.0 m/s−14.7 m/s=0.3 m/sv = 15.0\, \text{m/s} - 14.7\, \text{m/s} = 0.3\, \text{m/s}

(d) 2.00 seconds

At t=2.00t = 2.00 seconds, we can calculate the displacement and velocity of the ball using the equations of motion.

  • Displacement: y=0+(15.0 m/s)(2.00 s)+12(−9.8 m/s2)(2.00 s)2y = 0 + (15.0\, \text{m/s})(2.00\, \text{s}) + \frac{1}{2}(-9.8\, \text{m/s}^2)(2.00\, \text{s})^2 y=30.0 m−19.6 m=10.4 my = 30.0\, \text{m} - 19.6\, \text{m} = 10.4\, \text{m}
  • Velocity: v=15.0 m/s+(−9.8 m/s2)(2.00 s)v = 15.0\, \text{m/s} + (-9.8\, \text{m/s}^2)(2.00\, \text{s}) v=15.0 m/s−19.6 m/s=−4.6 m/sv = 15.0\, \text{m/s} - 19.6\, \text{m/s} = -4.6\, \text{m/s}

Conclusion

Q: What is the equation for calculating displacement?

A: The equation for calculating displacement is:

y=y0+v0t+12at2y = y_0 + v_0t + \frac{1}{2}at^2

where yy is the displacement, y0y_0 is the initial displacement, v0v_0 is the initial velocity, tt is time, and aa is acceleration.

Q: What is the equation for calculating velocity?

A: The equation for calculating velocity is:

v=v0+atv = v_0 + at

where vv is the velocity, v0v_0 is the initial velocity, aa is acceleration, and tt is time.

Q: What is the acceleration due to gravity?

A: The acceleration due to gravity is 9.8 m/s^2.

Q: How do I calculate the displacement and velocity of an object under constant acceleration?

A: To calculate the displacement and velocity of an object under constant acceleration, you can use the equations of motion:

  • Position: y=y0+v0t+12at2y = y_0 + v_0t + \frac{1}{2}at^2
  • Velocity: v=v0+atv = v_0 + at
  • Acceleration: a=−ga = -g (where gg is the acceleration due to gravity)

Q: What is the difference between displacement and velocity?

A: Displacement is the change in position of an object, while velocity is the rate of change of displacement.

Q: How do I calculate the displacement and velocity of an object thrown straight up?

A: To calculate the displacement and velocity of an object thrown straight up, you can use the equations of motion:

  • Position: y=y0+v0t+12(−g)t2y = y_0 + v_0t + \frac{1}{2}(-g)t^2
  • Velocity: v=v0+(−g)tv = v_0 + (-g)t

Q: What is the maximum height reached by an object thrown straight up?

A: The maximum height reached by an object thrown straight up is given by:

h=v022gh = \frac{v_0^2}{2g}

where hh is the maximum height, v0v_0 is the initial velocity, and gg is the acceleration due to gravity.

Q: How do I calculate the time it takes for an object to reach its maximum height?

A: To calculate the time it takes for an object to reach its maximum height, you can use the equation:

t=v0gt = \frac{v_0}{g}

where tt is the time, v0v_0 is the initial velocity, and gg is the acceleration due to gravity.

Q: What is the relationship between displacement, velocity, and acceleration?

A: The relationship between displacement, velocity, and acceleration is given by the equations of motion:

  • Position: y=y0+v0t+12at2y = y_0 + v_0t + \frac{1}{2}at^2
  • Velocity: v=v0+atv = v_0 + at
  • Acceleration: a=dvdta = \frac{dv}{dt}

These equations show that displacement, velocity, and acceleration are related to each other through the equations of motion.