A ball is thrown upward with an initial velocity \(V_0\) from the surface of the earth. The motion of the ball is affected by a drag force equal to \(m\gamma u^2\) (where m is mass of the ball, u is its instantaneous velocity and \(\gamma\) is a constant). Time taken by the ball to rise to its zenith is :
Show Hint
- \(\dfrac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\dfrac{\gamma}{g}}\,V_0\right)\)
- \(\dfrac{1}{\sqrt{\gamma g}}\ln\left(1+\sqrt{\dfrac{\gamma}{g}}\,V_0\right)\)
- \(\dfrac{1}{\sqrt{\gamma g}}\sin^{-1}\left(\sqrt{\dfrac{\gamma}{g}}\,V_0\right)\)
- \(\dfrac{1}{\sqrt{2\gamma g}}\tan^{-1}\left(\sqrt{\dfrac{2\gamma}{g}}\,V_0\right)\)
The Correct Option is A
Solution and Explanation
Time to reach the top (u = 0): \(t=\displaystyle\int_{0}^{V_0}\dfrac{du}{g+\gamma u^2}=\dfrac{1}{g}\int_0^{V_0}\dfrac{du}{1+(\gamma/g)u^2}\).
Using \(\displaystyle\int\dfrac{du}{1+k^2u^2}=\dfrac1k\tan^{-1}(ku)\) with \(k=\sqrt{\gamma/g}\):
\(t=\dfrac{1}{g}\sqrt{\dfrac{g}{\gamma}}\tan^{-1}\left(\sqrt{\dfrac{\gamma}{g}}V_0\right)=\dfrac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\dfrac{\gamma}{g}}V_0\right)\).
Answer: (1) \(\dfrac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\dfrac{\gamma}{g}}V_0\right)\)
Top JEE Main Physics Questions
- 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:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
Top JEE Main laws of motion Questions
- In the given arrangement of a doubly inclined plane, two blocks of masses \( M \) and \( m \) are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of \( m \), for which \( M = 10 \, \text{kg} \) will move down with an acceleration of \( 2 \, \text{m/s}^2 \), is: (take \( g = 10 \, \text{m/s}^2 \) and \( \tan 37^\circ = 3/4 \))
- Three blocks M1, M2, M3 having masses 4 kg, 6 kg and 10 kg respectively are hanging from a smooth pully using rope 1, 2 and 3 as shown in figure. The tension in the rope 1, T1 when they are moving upward with acceleration of 2ms–2 is ............... N (if g = 10 m/s2 )
- A light unstretchable string passing over a smooth light pulley connects two blocks of masses \(m_1\) and \(m_2\). If the acceleration of the system is \(\frac{g}{8}\), then the ratio of the masses \(\frac{m_2}{m_1}\) is:
- A cricket player catches a ball of mass 120 g moving with 25 m/s speed. If the catching process is completed in 0.1 s then the magnitude of force exerted by the ball on the hand of player will be (in SI unit):
The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature \( R = 2 \, \text{m} \). Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the side view mirror is \( a \). The value of \( 100a \) is _____________ m/s\(^2\).
Top JEE Main Questions
- 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 ______.
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,