Math Problem Statement

Please help me solve

A stone is thrown vertically upwards and falls back down. The motion is shown in the accompanying velocity-time graph. Solve for various aspects of the motion including acceleration, time to reach the peak, total time of flight, and maximum height.

Solution

Let's go through the solutions step by step.

Question 2: Motion of a stone thrown vertically

A stone is thrown upwards and falls back, following the velocity-time graph.

2.1 Acceleration of the stone

Since gravity is the only force acting on the stone (neglecting air resistance), the acceleration is constant and equal to g=9.8 m/s2g = 9.8 \text{ m/s}^2 downwards.

  • 2.1.1 From A to B: The stone is moving upwards but slowing down. The acceleration is 9.8 m/s2-9.8 \text{ m/s}^2 (downward).
  • 2.1.2 At B: At the peak, the velocity is zero, but acceleration is still 9.8 m/s2-9.8 \text{ m/s}^2 (downward).
  • 2.1.3 From B to C: The stone is moving downward, increasing in speed. The acceleration remains 9.8 m/s2-9.8 \text{ m/s}^2 (downward).

2.2 Value of t1t_1 (time to reach peak)

Using the first equation of motion: v=u+atv = u + at At the peak (point B), velocity is zero: 0=u+(9.8)t10 = u + (-9.8) t_1 From the graph, initial velocity u=7.84 m/su = 7.84 \text{ m/s}, so:

t1=7.849.8=0.8 st_1 = \frac{7.84}{9.8} = 0.8 \text{ s}

2.3 Value of t2t_2 (total time for the motion)

The motion is symmetric, meaning the time to go up equals the time to come down: t2=2t1=2(0.8)=1.6 st_2 = 2 t_1 = 2(0.8) = 1.6 \text{ s}

2.4 Value of x1x_1 (maximum height)

Using the kinematic equation: v2=u2+2asv^2 = u^2 + 2 a s Since velocity at peak is zero, we solve for ss (height):

0=(7.84)2+2(9.8)s0 = (7.84)^2 + 2(-9.8) s

s=(7.84)22(9.8)s = \frac{(7.84)^2}{2(9.8)}

s=61.465619.6=3.14 ms = \frac{61.4656}{19.6} = 3.14 \text{ m}


Question 3: Motion of a cricket ball

A cricket ball is thrown upwards at 14.7 m/s.

3.1 Calculate total time in the air

Using the equation: t1=ug=14.79.8=1.5 st_1 = \frac{u}{g} = \frac{14.7}{9.8} = 1.5 \text{ s}

Total time: t2=2t1=2(1.5)=3.0 st_2 = 2 t_1 = 2(1.5) = 3.0 \text{ s}

3.2 Velocity when returning to the pitcher’s hand

By symmetry, the velocity when it returns to the hand is the same magnitude but opposite direction to the initial velocity.

v=14.7 m/sv = -14.7 \text{ m/s}

3.3 Sketch the graphs

You need to draw:

  1. Position-Time Graph: A parabolic curve (reaching max height at 1.51.5 s, then coming down).
  2. Velocity-Time Graph: A straight line decreasing from 14.7 m/s to -14.7 m/s in 3 s.
  3. Acceleration-Time Graph: A constant horizontal line at 9.8 m/s2-9.8 \text{ m/s}^2.

Would you like any clarifications or more details? Here are some related questions to think about:

  1. What happens to acceleration when the stone reaches its peak?
  2. How does air resistance affect the motion of the stone?
  3. Why is the velocity-time graph a straight line?
  4. How would the motion change if the stone was thrown on another planet?
  5. Can we calculate the displacement using the area under the velocity-time graph?

Tip: The key to solving motion problems is to identify given values, use the correct equation, and check if motion is symmetrical!

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Math Problem Analysis

Mathematical Concepts

Kinematics
Motion
Velocity-Time Graph
Equations of Motion

Formulas

v = u + at
v^2 = u^2 + 2as
t = u/g

Theorems

Newton's Laws of Motion

Suitable Grade Level

Grades 9-12