3.8 Q-23

Question Statement

A ball is thrown vertically upward with an initial velocity of 2450 cm/s, neglecting air resistance. Find:

(i) The velocity of the ball at any time $t$.

(ii) The distance traveled by the ball in any time $t$.

(iii) The maximum height attained by the ball.

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Background and Explanation

This problem involves the motion of an object under the influence of gravity. We use Newton’s laws of motion to describe the motion of the ball. The acceleration due to gravity ($g$) is a constant value of $980 , \text{cm/s}^2$ (assuming downward direction). The velocity and position of the ball change over time based on this constant acceleration.

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Solution

(i) Velocity of the Ball at Any Time $t$

The rate of change of velocity is given by Newton’s second law of motion, which in this case states that the velocity changes due to gravity:

$$\frac{\mathrm{dv}}{\mathrm{dt}} = -g$$

Where $g = 980 , \text{cm/s}^2$ is the acceleration due to gravity. We can integrate both sides to find the velocity:

$$\int \mathrm{d}v = \int -g , \mathrm{d}t$$

Integrating:

$$v = -gt + C_1$$

Now, we apply the initial condition at $t = 0$, where the velocity is $2450 , \text{cm/s}$:

$$2450 = -g(0) + C_1$$ $$C_1 = 2450$$

Thus, the velocity equation becomes:

$$v = 2450 - 980t$$

This is the velocity of the ball at any time $t$.

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(ii) Distance Traveled in Any Time $t$

To find the distance traveled, we need to integrate the velocity equation. Recall that velocity is the derivative of position with respect to time:

$$v = \frac{\mathrm{dh}}{\mathrm{dt}} = 2450 - 980t$$

Integrating both sides:

$$\int \mathrm{dh} = \int (2450 - 980t) , \mathrm{dt}$$

This gives the position equation (height):

$$h = 2450t - 490t^2 + C_2$$

Applying the initial condition that the height is 0 when $t = 0$:

$$0 = 2450(0) - 490(0)^2 + C_2$$ $$C_2 = 0$$

Thus, the position equation becomes:

$$h = 2450t - 490t^2$$

This is the distance traveled by the ball at any time $t$.

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(iii) Maximum Height Attained by the Ball

At maximum height, the velocity of the ball becomes 0. We can use the velocity equation to find the time when the ball reaches maximum height:

$$0 = 2450 - 980t$$

Solving for $t$:

$$980t = 2450$$ $$t = \frac{2450}{980} = \frac{5}{2} , \text{seconds}$$

Now, substitute $t = \frac{5}{2}$ into the position equation to find the maximum height:

$$h_{\text{max}} = 2450 \times \frac{5}{2} - 490 \times \left( \frac{5}{2} \right)^2$$ $$h_{\text{max}} = 6125 - 3062.5$$ $$h_{\text{max}} = 3062.5 , \text{cm}$$

Thus, the maximum height attained by the ball is 3062.5 cm or 30.625 meters.

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Key Formulas or Methods Used

• Newton’s Law of Motion: $\frac{\mathrm{dv}}{\mathrm{dt}} = -g$
• Velocity equation: $v = -gt + C_1$
• Position equation: $h = 2450t - 490t^2$

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Summary of Steps

1. Velocity equation: Use $\frac{\mathrm{dv}}{\mathrm{dt}} = -g$ to find $v = 2450 - 980t$.
2. Position equation: Integrate the velocity equation to find $h = 2450t - 490t^2$.
3. Maximum height: Set $v = 0$ and solve for $t = \frac{5}{2}$ seconds.
4. Substitute $t$ into position equation to find $h_{\text{max}} = 3062.5 , \text{cm}$.