A boy throws a ball into air at $45^\circ$ from the horizontal to land it on a roof of a building of height $H$. If the ball attains maximum height in $2\,\text{s}$ and lands on the building in $3\,\text{s}$ after launch, then the value of $H$ is ___ m.
(Given: $g = 10\,\text{m s}^{-2}$)

Question

A particle of mass $m$ is projected with a velocity $u$ making an angle of $30^\circ$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is _____.

Answer 1

To solve this problem, we need to analyze the motion of the ball in two stages: ascending to the maximum height, and descending to the roof of the building.

Maximum Height Calculation:

We are given that the maximum height is reached in 2 seconds. For vertical motion, the time to reach maximum height $t_{max}$ is related to the initial vertical velocity $u_y$ by the formula:

$t_{max} = \frac{u_y}{g}$

Given $t_{max} = 2\, \text{s}$ and $g = 10\, \text{m/s}^2$, we find:

$u_y = g \times t_{max} = 10 \times 2 = 20\, \text{m/s}$

Horizontal and Vertical Components:

The launch angle is $45^\circ$, so the initial velocity vector is split equally into horizontal and vertical components:

$u_x = u_y = 20\, \text{m/s}$

Total Time of Flight:

The total time of flight given is 3 seconds, meaning the ball lands on the building 1 second after reaching its maximum height.

Vertical Displacement Calculation:

Using the equation of motion for the vertical displacement from the maximum height to the roof ($H$), we have:

$H = u_y \times t - \frac{1}{2} g \times t^2$

Where $t = 1\, \text{s}$ (time from max height to the roof):

$H = 0 \times 1 - \frac{1}{2} \times 10 \times 1^2 = -5\, \text{m}$ from the max height.

Since the ball reaches max height at $20\, \text{m}$ (calculation: $H_{max} = \frac{u_y^2}{2g} = \frac{20^2}{2 \times 10} = 20\, \text{m}$) from the launch point, at time $t=3s$ it is $20 - 5 = 15$ meters above the ground level.

Thus, the height of the building $H = 15\, \text{m}$.

The correct answer is $15$.

A boy throws a ball into air at $45^\circ$ from the horizontal to land it on a roof of a building of height $H$. If the ball attains maximum height in $2\,\text{s}$ and lands on the building in $3\,\text{s}$ after launch, then the value of $H$ is ___ m.
(Given: $g = 10\,\text{m s}^{-2}$)

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Projectile motion

Questions Asked in JEE Main exam

A boy throws a ball into air at $45^\circ$ from the horizontal to land it on a roof of a building of height $H$. If the ball attains maximum height in $2\,\text{s}$ and lands on the building in $3\,\text{s}$ after launch, then the value of $H$ is ___ m. (Given: $g = 10\,\text{m s}^{-2}$)

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Projectile motion

Questions Asked in JEE Main exam

A boy throws a ball into air at $45^\circ$ from the horizontal to land it on a roof of a building of height $H$. If the ball attains maximum height in $2\,\text{s}$ and lands on the building in $3\,\text{s}$ after launch, then the value of $H$ is ___ m.
(Given: $g = 10\,\text{m s}^{-2}$)