What is the formula for maximum height in projectile motion?

The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.

What is the formula for time of flight?

The total time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$.

How is time of flight related to maximum height?

By comparing the formulas, we find that $T$ is proportional to the square root of $H$ ($T \propto \sqrt{H}$).

Does the mass of the ball affect the time of flight?

No, in the absence of air resistance, the mass of the projectile does not affect its trajectory, height, or time of flight.

What is the ratio of heights given in the problem?

The problem states the first height is 8 times the second height, so $H_1/H_2 = 8$.

How do you calculate the ratio of $T_1/T_2$?

Since $T \propto \sqrt{H}$, the ratio is $\sqrt{H_1/H_2} = \sqrt{8} = 2\sqrt{2}$.

What happens if the initial velocities were different?

If initial velocities were different, the direct proportionality $T \propto \sqrt{H}$ would still hold as long as $g$ is constant.

Is the angle of projection necessary to solve this?

No, the relationship between $H$ and $T$ allows us to solve it without knowing the specific angles.

Why is the answer expressed as a ratio?

Ratios are standard in physics for comparing two similar scenarios with different parameters.

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. T1 and T2 are the total flying times of first and second ball, respectively, then the ratio of T1 and T2 is

Q: What is the value of $\sqrt{8}$?

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. T1 and T2 are the total flying times of first and second ball, respectively, then the ratio of T1 and T2 is | JEE Main Mathematics

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Q780MCQ

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. $T_1$ and $T_2$ are the total flying times of first and second ball, respectively, then the ratio of $T_1$ and $T_2$ is

✅ Correct Answer

2√2 : 1

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Solution Steps

Write formula for Maximum Height

For a projectile with initial velocity $u$ and angle $\theta$, the maximum height $H$ is:

$$H = \frac{u^2 \sin^2 \theta}{2g}$$

Write formula for Time of Flight

The total time of flight $T$ for the same projectile is:

$$T = \frac{2u \sin \theta}{g}$$

Establish Relationship between T and H

From the height formula, we can see that $\sin \theta = \frac{\sqrt{2gH}}{u}$. Substitute this into the time formula:

$$T = \frac{2u}{g} \left( \frac{\sqrt{2gH}}{u} \right) = 2\sqrt{\frac{2H}{g}}$$

Thus, $T \propto \sqrt{H}$.

Apply Given Conditions

It is given that $H_1 = 8 H_2$. Therefore, the ratio of times is:

$$\frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}}$$

Final Calculation

Substitute $H_1/H_2 = 8$ into the ratio:

$$\frac{T_1}{T_2} = \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$$

Final Ratio: $2\sqrt{2} : 1$

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Theory

1. Projectile Motion Components

Projectile motion is a form of motion experienced by an object or particle that is thrown near the Earth’s surface and moves along a curved path under the action of gravity only. It can be analyzed as two independent motions: a horizontal motion at constant velocity and a vertical motion with constant acceleration (gravity). The vertical component of velocity ($u \sin \theta$) determines both the maximum height and the time of flight.

2. Time of Flight (T)

The time of flight is the total time the projectile spends in the air. It is derived by considering the vertical displacement to be zero. The formula $T = 2u \sin \theta / g$ shows that the time depends directly on the initial vertical velocity component. Since gravity acts only vertically, the time it takes to reach the peak is exactly half of the total time of flight.

3. Maximum Height (H)

Maximum height is reached when the vertical component of velocity becomes zero ($v_y = 0$). Using the third equation of motion ($v^2 = u^2 + 2as$), we derive $H = (u \sin \theta)^2 / 2g$. This shows that the height reached is proportional to the square of the initial vertical velocity. Therefore, an object with twice the vertical speed will reach four times the height.

4. Interdependence of H and T

A very useful relation in projectile physics is $H = \frac{gT^2}{8}$. This derived formula shows that the maximum height is directly proportional to the square of the time of flight. If you know how much longer one ball stays in the air compared to another, you can immediately find the ratio of their heights without knowing their specific angles or velocities, provided they are on the same planet.

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FAQs

Does mass affect the ratio $T_1/T_2$?

No, mass does not appear in the kinematic equations for projectile motion.

What if air resistance is considered?

Air resistance would reduce both height and time, and the simple $\sqrt{H}$ relationship would no longer be exact.

Why is $T$ proportional to $\sqrt{H}$?

Because $H \propto \sin^2 \theta$ and $T \propto \sin \theta$. Taking the square root of $H$ gives a term proportional to $\sin \theta$.

Is the horizontal range related to height?

Yes, Range $R = 4H \cot \theta$.

Can we solve this using energy conservation?

Energy conservation helps find $H$ (potential energy at peak), but time is better handled using kinematic equations.

What is the angle for maximum $H$?

Maximum height is achieved when projected vertically upwards ($\theta = 90^\circ$).

What is the angle for maximum $T$?

Maximum time of flight also occurs at $\theta = 90^\circ$.

How do I simplify $\sqrt{8}$?

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$.

Does $g$ affect the ratio?

Since $g$ is constant for both balls, it cancels out during the ratio calculation.

If $H_1/H_2 = 16$, what would be $T_1/T_2$?

The ratio would be $\sqrt{16} = 4$.

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