(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

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

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3-2.Motion in Plane

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$A$ cricketer can throw a ball to a maximum horizontal distance of $100 \, m$. With the same effort,he throws the ball vertically upwards. The maximum height attained by the ball is ......... $m$.

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$A$ boy throws a ball upwards with velocity $v_0 = 20\, m/s$. The wind imparts a horizontal acceleration of $4\, m/s^2$ to the ball. The angle $\theta$ from the vertical at which the ball must be thrown so that the ball returns to the boy's hand is $(g = 10\, m/s^2)$

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The path of a projectile is given by the equation $y = ax - bx^2$, where $a$ and $b$ are constants, and $x$ and $y$ are the horizontal and vertical distances of the projectile from the point of projection, respectively. The maximum height attained by the projectile and the angle of projection are respectively:

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$A$ projectile is thrown at an angle of $\pi/3$ with the horizontal,reaching a maximum height $Y$. If another projectile is thrown with the same initial velocity at an angle of $\pi/6$,what will be the maximum height attained?

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$A$ ball is projected from the ground with a velocity $V$ at an angle $\theta$ to the vertical. On its path,it makes an elastic collision with a vertical wall and returns to the ground. The total time of flight of the ball is:

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$A$ cricketer can throw a ball to a maximum horizontal distance of $100 \, m$. With the same effort,he throws the ball vertically upwards. The maximum height attained by the ball is ......... $m$.

$A$ boy throws a ball upwards with velocity $v_0 = 20\, m/s$. The wind imparts a horizontal acceleration of $4\, m/s^2$ to the ball. The angle $\theta$ from the vertical at which the ball must be thrown so that the ball returns to the boy's hand is $(g = 10\, m/s^2)$

$A$ projectile is thrown at an angle of $\pi/3$ with the horizontal,reaching a maximum height $Y$. If another projectile is thrown with the same initial velocity at an angle of $\pi/6$,what will be the maximum height attained?

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