Math Problem Statement

A ball is thrown straight upward with an initial velocity vi and in a time t reaches the top of its flight path, which is a displacement yf above the launch point. Consider that same ball being thrown upward with an initial velocity of 2 x vi x a. Determine the displacement of the top point above the launch path.

Solution

To determine the displacement of the ball's top point when thrown with an initial velocity of 2×vi×a2 \times v_i \times a, let's break the problem down step by step using kinematic equations.

Step 1: Displacement for the initial velocity viv_i

We know that the ball is thrown upward with an initial velocity viv_i, and after time tt, it reaches the top of its flight path where its velocity becomes zero. The displacement from the launch point to the top of the flight path is given by yfy_f.

Using the kinematic equation:

vf2=vi2+2aΔyv_f^2 = v_i^2 + 2a \Delta y

Where:

  • vf=0v_f = 0 (at the top of the flight path),
  • viv_i is the initial velocity,
  • aa is the acceleration due to gravity, g-g,
  • Δy=yf\Delta y = y_f is the displacement.

Substitute vf=0v_f = 0 and a=ga = -g:

0=vi22gyf0 = v_i^2 - 2g y_f

Solving for yfy_f:

yf=vi22gy_f = \frac{v_i^2}{2g}

Step 2: Displacement for the new initial velocity 2×vi×a2 \times v_i \times a

Now, the ball is thrown with a new initial velocity of 2×vi×a2 \times v_i \times a.

Using the same kinematic equation for the new situation:

vf2=(2via)2+2aΔyv_f^2 = (2 v_i a)^2 + 2a \Delta y'

Substitute vf=0v_f = 0 and a=ga = -g:

0=(2via)22gyf0 = (2 v_i a)^2 - 2g y_f'

Solving for yfy_f', the displacement for the new velocity:

yf=(2via)22gy_f' = \frac{(2 v_i a)^2}{2g}

yf=4vi2a22gy_f' = \frac{4 v_i^2 a^2}{2g}

Thus, the new displacement to the top point of the flight path is:

yf=2a2×vi2gy_f' = 2 a^2 \times \frac{v_i^2}{g}

Final Answer:

The displacement of the top point above the launch path when the ball is thrown with an initial velocity of 2×vi×a2 \times v_i \times a is:

yf=2a2×yfy_f' = 2 a^2 \times y_f

This means the new displacement is scaled by 2a22a^2 times the original displacement yfy_f.


Would you like more details or have any questions?

Here are 5 related questions:

  1. What happens to the flight time when the initial velocity is doubled?
  2. How does the maximum height change if the initial velocity is tripled?
  3. What is the effect of air resistance on the displacement in such a problem?
  4. How would you calculate the time to reach the top with the new velocity?
  5. How can you express the velocity at any point in the trajectory?

Tip: When solving kinematic problems, always check which kinematic equation fits the known quantities like initial velocity, time, and acceleration!

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Math Problem Analysis

Mathematical Concepts

Kinematics
Projectile Motion

Formulas

vf^2 = vi^2 + 2aΔy
y_f = vi^2 / 2g
y_f' = (2 vi a)^2 / 2g

Theorems

Kinematic Equations

Suitable Grade Level

Grades 9-12