Question:
\(y=(Pt^2-Qt^3)\;m\) is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is
\(y=(Pt^2-Qt^3)\;m\) is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is
Show Hint
For maximum height in vertical motion, differentiate displacement with respect to time and put velocity equal to zero.
Updated On: Jun 22, 2026
- \(\dfrac{27P^3}{4Q^2}\)
- \(\dfrac{4Q^2}{27P^3}\)
- \(\dfrac{4P^3}{27Q^2}\)
- \(\dfrac{27Q^2}{4P^3}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Write the given displacement equation.
\[ y=Pt^2-Qt^3 \] For maximum height, velocity in the vertical direction must become zero.
So, \[ \frac{dy}{dt}=0 \]
Step 2: Differentiate \(y\) with respect to \(t\).
\[ \frac{dy}{dt}=2Pt-3Qt^2 \] \[ \frac{dy}{dt}=t(2P-3Qt) \] For maximum height, \[ t(2P-3Qt)=0 \] Thus, \[ t=0 \] or \[ 2P-3Qt=0 \] \[ t=\frac{2P}{3Q} \] The non-zero time at which maximum height is reached is \[ t=\frac{2P}{3Q} \]
Step 3: Substitute this value of \(t\) in \(y\).
\[ y_{\max}=P\left(\frac{2P}{3Q}\right)^2-Q\left(\frac{2P}{3Q}\right)^3 \] \[ =P\left(\frac{4P^2}{9Q^2}\right)-Q\left(\frac{8P^3}{27Q^3}\right) \] \[ =\frac{4P^3}{9Q^2}-\frac{8P^3}{27Q^2} \] Taking LCM, \[ y_{\max}=\frac{12P^3-8P^3}{27Q^2} \] \[ y_{\max}=\frac{4P^3}{27Q^2} \]
Step 4: Final conclusion.
Hence, the maximum height reached by the ball is \[ \boxed{\frac{4P^3}{27Q^2}} \]
\[ y=Pt^2-Qt^3 \] For maximum height, velocity in the vertical direction must become zero.
So, \[ \frac{dy}{dt}=0 \]
Step 2: Differentiate \(y\) with respect to \(t\).
\[ \frac{dy}{dt}=2Pt-3Qt^2 \] \[ \frac{dy}{dt}=t(2P-3Qt) \] For maximum height, \[ t(2P-3Qt)=0 \] Thus, \[ t=0 \] or \[ 2P-3Qt=0 \] \[ t=\frac{2P}{3Q} \] The non-zero time at which maximum height is reached is \[ t=\frac{2P}{3Q} \]
Step 3: Substitute this value of \(t\) in \(y\).
\[ y_{\max}=P\left(\frac{2P}{3Q}\right)^2-Q\left(\frac{2P}{3Q}\right)^3 \] \[ =P\left(\frac{4P^2}{9Q^2}\right)-Q\left(\frac{8P^3}{27Q^3}\right) \] \[ =\frac{4P^3}{9Q^2}-\frac{8P^3}{27Q^2} \] Taking LCM, \[ y_{\max}=\frac{12P^3-8P^3}{27Q^2} \] \[ y_{\max}=\frac{4P^3}{27Q^2} \]
Step 4: Final conclusion.
Hence, the maximum height reached by the ball is \[ \boxed{\frac{4P^3}{27Q^2}} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Motion in a straight line Questions
- A juggler throws balls vertically into the air such that he throws the next ball when the previous one is at its highest point. If he throws 3 balls each second, then the maximum height reached by each ball is
(Acceleration due to gravity \(g = 10~\text{ms}^{-2}\)) - A particle under constant acceleration covers the first half of the total distance in time $t_1$ and remaining in time $t_2$. Then:
- A particle moves with velocity $v = at - bt^2$ where a and b are constants. The acceleration becomes zero at:
- A ball is dropped from a tower of height \(80\,\text{m}\). The time it takes to cover the last \(50\%\) of its fall is (acceleration due to gravity \(=10\,\text{m s}^{-2}\))
- A passenger sees a bus standing 10 m ahead of him. The bus starts moving with a constant acceleration of \(5 \, m\,s^{-2}\) away from him. The passenger runs with constant speed to catch the bus. The minimum speed required to catch the bus is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions