$A$ particle $(A)$ is dropped from a height and another particle $(B)$ is thrown in a horizontal direction with a speed of $5 \; m/s$ from the same height. The correct statement is:
- ABoth particles will reach the ground simultaneously.
- BBoth particles will reach the ground with the same speed.
- CParticle $(A)$ will reach the ground first with respect to particle $(B)$.
- DParticle $(B)$ will reach the ground first with respect to particle $(A)$.
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$A$ body is projected horizontally from the top of a tower of height $180 \,m$ with a velocity of $20 \,ms^{-1}$. If acceleration due to gravity is $10 \,ms^{-2}$, then match the following:
$A$. Velocity of the body after $1 \,s$ (in $ms^{-1}$)
$I$. $5$
$B$. Horizontal displacement of the body after $1 \,s$ (in $m$)
$II$. $20$
$C$. Vertical displacement of the body after $1 \,s$ (in $m$)
$III$. $10$
$D$. Vertical velocity of the body after $1 \,s$ (in $ms^{-1}$)
$IV$. $22.4$
| $A$. Velocity of the body after $1 \,s$ (in $ms^{-1}$) | $I$. $5$ |
| $B$. Horizontal displacement of the body after $1 \,s$ (in $m$) | $II$. $20$ |
| $C$. Vertical displacement of the body after $1 \,s$ (in $m$) | $III$. $10$ |
| $D$. Vertical velocity of the body after $1 \,s$ (in $ms^{-1}$) | $IV$. $22.4$ |
The equations of motion of a projectile are given by $x = 36t$ metre and $2y = 96t - 9.8t^2$ metre. The angle of projection is:
$A$ projectile is thrown with a velocity of $10 \sqrt{2} \ m/s$ at an angle of $45^{\circ}$ with the horizontal. The time interval between the moments when the speed is $\sqrt{125} \ m/s$ is $(g = 10 \ m/s^2)$. (in $s$)
$A$ body is projected horizontally from the top of a tall tower with a velocity of $30 \,ms^{-1}$. At time $t_1$, its horizontal and vertical components of the velocity are equal and at time $t_2$, its horizontal and vertical displacements are equal. Then $t_2-t_1$ is (take, $g=10 \,ms^{-2}$ ) (in $\,s$)
$A$ particle is projected from the ground with an initial speed of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and the time it reaches the highest point of its trajectory is
DifficultAP EAMCET 2013
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