A body 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 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 man wants to reach from $A$ to the opposite corner of the square $C$. The sides of the square are $100\, m$. A central square of $50\, m\,\times \,50\, m$ is filled with sand. Outside this square, he can walk at a speed $1\,ms^{-1}$. In the central square, he can walk only at a speed of $v\,ms^{-1}$ $(v < 1)$. What is smallest value of $v$ for which he can reach faster via a straight path through the sand than any path in the square outside the sand ?
A particle starts from the origin at $\mathrm{t}=0$ with an initial velocity of $3.0 \hat{\mathrm{i}} \;\mathrm{m} / \mathrm{s}$ and moves in the $x-y$ plane with a constant acceleration $(6.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}) \;\mathrm{m} / \mathrm{s}^{2} .$ The $\mathrm{x}$ -coordinate of the particle at the instant when its $y-$coordinate is $32\;\mathrm{m}$ is $D$ meters. The value of $D$ is
A person walks $25.0^{\circ}$ north of east for $3.18 \,km$. How far would she have to walk due north and then due east to arrive at the same location?