Motion in a Plane
19 Board Physics previous year questions on Motion in a Plane — options free on every question; 2 include the answer & explanation free, the rest unlock with PYQ Pass.
A particle is projected with a velocity \(60\text{ }\text{m/s}\) at an angle \(3{0}^{∘}\) with respect to the horizontal. It reaches a height \({h}_{1}\) in the first second and height \({h}_{2}\) in the last second during its upward motion. Find the ratio of \(\frac{{h}_{1}}{{h}_{2}}\). (JEE Mains - 21 Jan 2025 - Shift I Memory Based)
5:1
-
Initial Vertical Velocity (\({u}_{y}\)):
\({u}_{y}=u\sin \theta =60\sin 3{0}^{∘}=30\text{ }\text{m/s}\mathrm{.}\) -
Time of Flight (Total time of upward motion):
\({T}_{\text{up}}=\frac{{u}_{y}}{g}=\frac{30}{10}=3\text{ }\text{seconds}\mathrm{.}\)
Using the formula for the total time of motion upward:The last second in the upward motion corresponds to the time \(t=2\text{ }\text{seconds}\to 3\text{ }\text{seconds}\)
Using the equation of motion:
\(h={u}_{y}t−\frac{1}{2}g{t}^{2},\)
for \(t=1\):
\({h}_{1}=(30)(1)−\frac{1}{2}(10)(1{)}^{2}=30−5=25\text{ }\text{m}\mathrm{.}\)
Height in the Last Second Upward (\({h}_{2}\)):For the last second in the upward motion, the time interval is \(t=2\to 3\text{ }\text{seconds}\). To find \({h}_{2}\), calculate the difference in height between these two time intervals:
\({h}_{2}=\text{Height at }t=3\text{ }\text{seconds}−\text{Height at }t=2\text{ }\text{seconds}\mathrm{.}\)
Height at \(t=2\):\(h(t=2)=(30)(2)−\frac{1}{2}(10)(2{)}^{2}=60−20=40\text{ }\text{m}\mathrm{.}\)
Height at \(t=3\):\(h(t=3)=(30)(3)−\frac{1}{2}(10)(3{)}^{2}=90−45=45\text{ }\text{m}\mathrm{.}\)
Height in the last second:\({h}_{2}=h(t=3)−h(t=2)=45−40=5\text{ }\text{m}\mathrm{.}\)
Ratio \(\frac{{h}_{1}}{{h}_{2}}\):\(\frac{{h}_{1}}{{h}_{2}}=\frac{25}{5}=5:1.\)
Final Answer:
\(5:1\)
If the two projectile are projected with the same speed and the angles of projection for two projectiles are 30° and 60° then the ratio of velocities at maximum height is:
\(\sqrt{3}:1\)
The maximum height of the projectile is given by
\(\mathrm{H}=\frac{{\mathrm{u}}^{2}{\sin }^{2}\theta }{2\mathrm{g}}\\ \mathrm{So},\\ \frac{{\mathrm{H}}_{1}}{{\mathrm{H}}_{2}}=\frac{{\sin }^{2}{\theta }_{1}}{{\sin }^{2}{\theta }_{2}}=\frac{{\sin }^{2}60^\circ }{{\sin }^{2}30^\circ }\\ \frac{{\mathrm{H}}_{1}}{{\mathrm{H}}_{2}}=\sqrt{3}:1\)
If the radius of curvature of the path of two particles of same mass are in the ratio 3:4, then in order to have constant centripetal force, their velocities will be in the ratio of:
Options are free to see. Unlock the correct answer and full explanation with Pass.
The angle of projection for a projectile to have same horizontal range and maximum height is
[JEE Main 2024, 8 Apr (Shift 2)]:
Options are free to see. Unlock the correct answer and full explanation with Pass.
Two projectiles are fired with same initial speed from same point on ground at angles of \(\left(45^\circ -\alpha \right)\) and \(\left(45^\circ +\alpha \right)\), respectively, with the horizontal direction. The ratio of their maximum heights attained is :
[JEE Main 2025, 29 Jan (Shift 1)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
Two particles are projected with the same velocity but at different projection angles: \(\left(\frac{\pi }{4}+\alpha \right)\) and \(\left(\frac{\pi }{4}-\alpha \right)\). Determine the ratio of their maximum heights.
Options are free to see. Unlock the correct answer and full explanation with Pass.
A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle \(\theta\) to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection \(\theta\) is then given by:
Options are free to see. Unlock the correct answer and full explanation with Pass.
The angle of projection for a projectile to have same horizontal range and maximum height is
[JEE Main 2024, 8 Apr (Shift 2)]:
Options are free to see. Unlock the correct answer and full explanation with Pass.
If the angles of projection for two projectiles are \(3{0}^{∘}\) and \(6{0}^{∘}\), what is the ratio of their velocities at maximum height?
Options are free to see. Unlock the correct answer and full explanation with Pass.
Two projectiles are fired with same initial speed from same point on ground at angles of \(\left(45^\circ -\alpha \right)\) and \(\left(45^\circ +\alpha \right)\), respectively, with the horizontal direction. The ratio of their maximum heights attained is :
[JEE Main 2025, 29 Jan (Shift 1)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
If the radius of curvature of the path of two particles of same mass are in the ratio 3:4, then in order to have constant centripetal force, their velocities will be in the ratio of:
Options are free to see. Unlock the correct answer and full explanation with Pass.
Two particles are projected with the same velocity but at different projection angles: \(\left(\frac{\pi }{4}+\alpha \right)\) and \(\left(\frac{\pi }{4}-\alpha \right)\). Determine the ratio of their maximum heights.
Options are free to see. Unlock the correct answer and full explanation with Pass.
The position vector of a moving body at any instant of time is given as \(\vec{r}=\left(5{t}^{2}\hat{ı}-5t\hat{ȷ}\right)m\). The magnitude and direction of velocity at t = 2 s is,
[JEE Main 2025, 24 Jan (Shift 2)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
If the radius of curvature of the path of two particles of same mass are in the ratio 3:4, then in order to have constant centripetal force, their velocities will be in the ratio of:
Options are free to see. Unlock the correct answer and full explanation with Pass.
The position vector of a moving body at any instant of time is given as \(\vec{r}=\left(5{t}^{2}\hat{ı}-5t\hat{ȷ}\right)m\). The magnitude and direction of velocity at t = 2 s is,
[JEE Main 2025, 24 Jan (Shift 2)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
The angle of projection for a projectile to have same horizontal range and maximum height is :
[JEE Main 2024, 08 Apr (Shift 2)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
If the radius of curvature of the path of two particles of same mass are in the ratio 3:4, then in order to have constant centripetal force, their velocities will be in the ratio of:
Options are free to see. Unlock the correct answer and full explanation with Pass.
A ball of mass 100 g is projected with velocity \(20\mathrm{m}/\mathrm{s}\) at \(60^\circ\) with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is
[JEE Main 2025, 22 Jan (Shift 2)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
The initial speed of a projectile fired from ground is u. At the highest point during its motion, the speed of projectile is \(\frac{\sqrt{3}}{2}u\). The time of flight of the projectile is:
Options are free to see. Unlock the correct answer and full explanation with Pass.
Practice more Board Physics PYQs
Browse every Physics chapter, or explore the full Board question bank.
All Physics chapters →