(B) The time of flight $T$ for the ball thrown vertically upward with speed $u$ is given by $T = \frac{2u}{g}$.Since the truck is moving with a consta

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(B) The time of flight $T$ for the ball thrown vertically upward with speed $u$ is given by $T = \frac{2u}{g}$.Since the truck is moving with a constant horizontal speed $v$ and the ball retains this horizontal velocity component,the ball will return to the same point on the truck.The distance $d$ traversed by the truck during this time $T$ is given by $d = v \times T$.Substituting the value of $T$,we get $d = v \times \frac{2u}{g} = \frac{2uv}{g}$.

Class 11 Physics 2 Motion in a Plane Mix Examples-Motion in Plane

Medium

$A$ truck is moving on a horizontal road with a constant speed $v$. $A$ ball is thrown from the truck vertically upward at a speed $u$ with respect to the truck. What is the distance traversed by the truck when the ball returns to the truck?

A
$\frac{uv}{g}$
B
$\frac{2uv}{g}$
C
$\frac{3uv}{g}$
D
$\frac{uv}{2g}$

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2 Motion in a Plane

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Mix Examples-Motion in Plane

$A$ projectile has the same range $R$ for two angles of projection. If $T_1$ and $T_2$ are the times of flight in the two cases,then $R$ is

Medium

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$A$ bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \; km$ away. By adjusting its angle of projection,can one hope to hit a target $5.0 \; km$ away? Assume the muzzle speed to be fixed,and neglect air resistance.

Medium

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$A$ particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle $\theta$ to the horizontal,the maximum height attained by it equals $4R$. The angle of projection,$\theta$,is then given by

MediumNEET 2021

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$A$ ball of mass $1 \; kg$ is thrown vertically upwards and returns to the ground after $3 \; s$. Another ball,thrown at $60^{\circ}$ with the vertical,also stays in the air for the same time before it touches the ground. The ratio of the two maximum heights reached is:

MediumNEET 2017

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$A$ projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20 \; m/s$. After $10 \; s$,its inclination with the horizontal is $\beta$. The value of $\tan \beta$ will be: $(g = 10 \; m/s^2)$

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$A$ truck is moving on a horizontal road with a constant speed $v$. $A$ ball is thrown from the truck vertically upward at a speed $u$ with respect to the truck. What is the distance traversed by the truck when the ball returns to the truck?

Similar Questions

$A$ projectile has the same range $R$ for two angles of projection. If $T_1$ and $T_2$ are the times of flight in the two cases,then $R$ is

$A$ bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \; km$ away. By adjusting its angle of projection,can one hope to hit a target $5.0 \; km$ away? Assume the muzzle speed to be fixed,and neglect air resistance.

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