Question:
A sphere rolls down from the top of an inclined plane which makes an angle \(30^\circ\) with the horizontal roof. If the highest and lowest points of the inclined plane are \(8.75\,\text{m}\) and \(3.75\,\text{m}\) respectively from the ground, then the horizontal distance from the lower edge of the roof at which the sphere hits the ground is \((g=10\,\text{m/s}^2)\):
A sphere rolls down from the top of an inclined plane which makes an angle \(30^\circ\) with the horizontal roof. If the highest and lowest points of the inclined plane are \(8.75\,\text{m}\) and \(3.75\,\text{m}\) respectively from the ground, then the horizontal distance from the lower edge of the roof at which the sphere hits the ground is \((g=10\,\text{m/s}^2)\):
Show Hint
For projectile motion from a height, first find the velocity at the point of projection, then resolve it into horizontal and vertical components.
Updated On: Jun 24, 2026
- \(5\,\text{m}\)
- \(5\sqrt{3}\,\text{m}\)
- \(\dfrac{5\sqrt{3}}{2}\,\text{m}\)
- \(10\,\text{m}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Find the vertical fall on the inclined plane.
Highest point height: \[ 8.75\,\text{m} \] Lowest point height: \[ 3.75\,\text{m} \] So, vertical fall is \[ 8.75-3.75=5\,\text{m} \]
Step 2: Find the velocity at the lower edge.
Using conservation of energy, \[ mgh=\frac{1}{2}mv^2 \] \[ v^2=2gh \] \[ v^2=2(10)(5)=100 \] \[ v=10\,\text{m/s} \]
Step 3: Resolve velocity components.
The sphere leaves the roof at an angle \(30^\circ\) below the horizontal.
Horizontal component: \[ u_x=10\cos30^\circ=10\cdot \frac{\sqrt{3}}{2}=5\sqrt{3} \] Vertical downward component: \[ u_y=10\sin30^\circ=10\cdot \frac{1}{2}=5 \]
Step 4: Find time to reach the ground.
The sphere falls from height \[ 3.75\,\text{m} \] Using \[ s=u_yt+\frac{1}{2}gt^2 \] \[ 3.75=5t+5t^2 \] \[ 5t^2+5t-3.75=0 \] Solving, \[ t=\frac{1}{2}\,\text{s} \]
Step 5: Find horizontal distance.
Horizontal distance is \[ R=u_xt \] \[ R=5\sqrt{3}\times \frac{1}{2} \] \[ R=\frac{5\sqrt{3}}{2}\,\text{m} \]
Step 6: Final conclusion.
Hence, \[ \boxed{\frac{5\sqrt{3}}{2}\,\text{m}} \]
Highest point height: \[ 8.75\,\text{m} \] Lowest point height: \[ 3.75\,\text{m} \] So, vertical fall is \[ 8.75-3.75=5\,\text{m} \]
Step 2: Find the velocity at the lower edge.
Using conservation of energy, \[ mgh=\frac{1}{2}mv^2 \] \[ v^2=2gh \] \[ v^2=2(10)(5)=100 \] \[ v=10\,\text{m/s} \]
Step 3: Resolve velocity components.
The sphere leaves the roof at an angle \(30^\circ\) below the horizontal.
Horizontal component: \[ u_x=10\cos30^\circ=10\cdot \frac{\sqrt{3}}{2}=5\sqrt{3} \] Vertical downward component: \[ u_y=10\sin30^\circ=10\cdot \frac{1}{2}=5 \]
Step 4: Find time to reach the ground.
The sphere falls from height \[ 3.75\,\text{m} \] Using \[ s=u_yt+\frac{1}{2}gt^2 \] \[ 3.75=5t+5t^2 \] \[ 5t^2+5t-3.75=0 \] Solving, \[ t=\frac{1}{2}\,\text{s} \]
Step 5: Find horizontal distance.
Horizontal distance is \[ R=u_xt \] \[ R=5\sqrt{3}\times \frac{1}{2} \] \[ R=\frac{5\sqrt{3}}{2}\,\text{m} \]
Step 6: Final conclusion.
Hence, \[ \boxed{\frac{5\sqrt{3}}{2}\,\text{m}} \]
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 Rotational motion Questions
- 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 $Q$ )
- A planet is revolving around the Sun as shown in the figure. The radius vectors joining the Sun and the planet at points $A$ and $B$ are $90 \times 10^{6} km$ and $60 \times 10^{6} km$, respectively. The ratio of velocities of the planet at the points $A$ and $B$ when its velocities make angle $30^{\circ}$ and $60^{\circ}$ with major-axis of the orbit is
- The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum $ (\text{in kg}\,{{m}^{2}}{{s}^{-1}}) $ is approximately:
- The moment of inertia of a solid sphere of mass 20 kg and diameter 20 cm about the tangent to the sphere is:
- A disc starts rotating from rest with constant acceleration and attains angular velocity of 20 rad/s in 5 seconds. The total angular displacement during this interval 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