(C) The velocity components are obtained by differentiating the position equations with respect to time $t$.$v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) =

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(C) The velocity components are obtained by differentiating the position equations with respect to time $t$.$v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) = 6 \ m/s$.$v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5t^2) = 8 - 10t \ m/s$.At the time of projection,$t = 0$.Substituting $t = 0$ into the velocity components:$v_x = 6 \ m/s$.$v_y = 8 - 10(0) = 8 \ m/s$.The magnitude of the initial velocity $v$ is given by $v = \sqrt{v_x^2 + v_y^2}$.$v = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ m/s$.

Class 11 Physics 2 Motion in a Plane Horizontal Projectile Motion

Medium

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The velocity with which the projectile is projected is ......... $m/s$.

A
$8$
B
$6$
C
$10$
D
Not obtainable from the data

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Class 11

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2 Motion in a Plane

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Horizontal Projectile Motion

$A$ ball is projected with kinetic energy $E$, at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will be:

MediumJEE MAIN 2022

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$A$ particle is projected horizontally from a tower with velocity $10\,m/s$. Taking $g=10\,m/s^2$,match the following two columns at time $t=1\,s$.
Column $I$ Column $II$
$(A)$ Horizontal component of velocity $(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity $(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement $(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement $(s)$ $20$ $SI$ unit

Medium

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$A$ particle covers $50\, m$ distance when projected with an initial speed. On the same surface,it will cover a distance of ......... $m$ when projected with double the initial speed.

Easy

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$A$ particle is projected in air at some angle to the horizontal and moves along a parabola as shown in the figure,where $x$ and $y$ indicate horizontal and vertical directions,respectively. The diagram shows the direction of velocity and acceleration at points $A$,$B$,and $C$.

Medium

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$A$ projectile is fired at an angle of $45^{\circ}$ with the horizontal. The elevation angle of the projectile at its highest point as seen from the point of projection is:

EasyTS EAMCET 2020

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Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The velocity with which the projectile is projected is ......... $m/s$.

Similar Questions

$A$ ball is projected with kinetic energy $E$, at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will be:

$A$ particle covers $50\, m$ distance when projected with an initial speed. On the same surface,it will cover a distance of ......... $m$ when projected with double the initial speed.

$A$ particle is projected in air at some angle to the horizontal and moves along a parabola as shown in the figure,where $x$ and $y$ indicate horizontal and vertical directions,respectively. The diagram shows the direction of velocity and acceleration at points $A$,$B$,and $C$.

$A$ projectile is fired at an angle of $45^{\circ}$ with the horizontal. The elevation angle of the projectile at its highest point as seen from the point of projection is:

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