Question:
Two identical bodies, projected with the same speed at two different angles cover the same horizontal range \(R\). If the time of flight of these bodies are 5 s and 10 s, respectively, then the value of \(R\) is ________ m. (Take \(g = 10\) m/s²)
Two identical bodies, projected with the same speed at two different angles cover the same horizontal range \(R\). If the time of flight of these bodies are 5 s and 10 s, respectively, then the value of \(R\) is ________ m. (Take \(g = 10\) m/s²)
Updated On: Apr 13, 2026
- 250
- 25
- 500
- 125
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Two projectiles have the same range for the same speed if their angles of projection are complementary (\(\theta\) and \(90^\circ - \theta\)). We can relate the range to the product of their times of flight.
Step 2: Key Formula or Approach:
1. Time of flight: \(T_1 = \frac{2u \sin \theta}{g}\) and \(T_2 = \frac{2u \cos \theta}{g}\).
2. Horizontal Range: \(R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}\).
Step 3: Detailed Explanation:
Multiply the two times of flight: \[ T_1 T_2 = \left(\frac{2u \sin \theta}{g}\right) \left(\frac{2u \cos \theta}{g}\right) = \frac{2}{g} \left(\frac{2u^2 \sin \theta \cos \theta}{g}\right) \] Notice the term in the parentheses is the formula for Range \(R\): \[ T_1 T_2 = \frac{2R}{g} \] Rearranging for \(R\): \[ R = \frac{g T_1 T_2}{2} \] Substitute the given values (\(g = 10, T_1 = 5, T_2 = 10\)): \[ R = \frac{10 \times 5 \times 10}{2} = \frac{500}{2} = 250 \text{ m} \]
Step 4: Final Answer:
The horizontal range \(R\) is 250 m.
Two projectiles have the same range for the same speed if their angles of projection are complementary (\(\theta\) and \(90^\circ - \theta\)). We can relate the range to the product of their times of flight.
Step 2: Key Formula or Approach:
1. Time of flight: \(T_1 = \frac{2u \sin \theta}{g}\) and \(T_2 = \frac{2u \cos \theta}{g}\).
2. Horizontal Range: \(R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}\).
Step 3: Detailed Explanation:
Multiply the two times of flight: \[ T_1 T_2 = \left(\frac{2u \sin \theta}{g}\right) \left(\frac{2u \cos \theta}{g}\right) = \frac{2}{g} \left(\frac{2u^2 \sin \theta \cos \theta}{g}\right) \] Notice the term in the parentheses is the formula for Range \(R\): \[ T_1 T_2 = \frac{2R}{g} \] Rearranging for \(R\): \[ R = \frac{g T_1 T_2}{2} \] Substitute the given values (\(g = 10, T_1 = 5, T_2 = 10\)): \[ R = \frac{10 \times 5 \times 10}{2} = \frac{500}{2} = 250 \text{ m} \]
Step 4: Final Answer:
The horizontal range \(R\) is 250 m.
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
View More Questions
Top JEE Main Kinetic Energy and Projectile Motion Questions
- A ball having kinetic energy \( KE \), is projected at an angle of \( 60^\circ \) from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?
- A ball of mass 100 g is projected with velocity 20 m/s at \( 60^\circ \) with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions