(D) The time taken to reach the maximum height is given by $t_m = \frac{u_y}{g} = 2 \text{ s}$.Given $g = 10 \text{ m/s}^2$, we have $u_y = 2 \times 1

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(D) The time taken to reach the maximum height is given by $t_m = \frac{u_y}{g} = 2 \text{ s}$.Given $g = 10 \text{ m/s}^2$, we have $u_y = 2 \times 10 = 20 \text{ m/s}$.The vertical displacement $y$ at any time $t$ is given by the equation $y = u_y t - \frac{1}{2} g t^2$.At $t = 3 \text{ s}$, the ball is at height $H$, so:$H = (20 \text{ m/s}) \times (3 \text{ s}) - \frac{1}{2} \times (10 \text{ m/s}^2) \times (3 \text{ s})^2$$H = 60 \text{ m} - 5 \times 9 \text{ m}$$H = 60 \text{ m} - 45 \text{ m} = 15 \text{ m}$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

Medium

$A$ boy throws a ball into the air at $45^{\circ}$ from the horizontal to land it on the 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 . . . . . . $\text{m}$. $(g = 10 \text{ m/s}^2)$

JEE MAIN 2026 All JEE MAIN

A
$20$
B
$10$
C
$25$
D
$15$

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3-2.Motion in Plane

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Horizontal Projectile Motion

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At any instant $t$, the vertical distance $Y$ and horizontal distance $X$ of a projectile are given by $2Y = 6t - gt^2$ and $X = 4t$. The initial velocity of the projectile is ($X$ and $Y$ are in $m$ and $t$ is in $s$). (in $\,m/s$)

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At any instant $t$, the vertical distance $Y$ and horizontal distance $X$ of a projectile are given by $2Y = 6t - gt^2$ and $X = 4t$. The initial velocity of the projectile is ($X$ and $Y$ are in $m$ and $t$ is in $s$). (in $\,m/s$)

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