A body is projected horizontally from the top of a tower of height...

Given, initial horizontal component of velocity, $u_x=20 \mathrm{~m} / \mathrm{s} \text{. }$ Initial vertical component of velocity, $u_y=0$ Acceleration due to gravity, $g=a_y=10 \mathrm{~m} / \mathrm{s}^2$ So, horizontal component of velocity after $1 \mathrm{~s}$, $$\begin{aligned} v_x & =u_x+a_x t \\ & =20+0=20 \mathrm{~m} / \mathrm{s} \end{aligned}$$ and vertical component of velocity after $1 \mathrm{~s}$, $$\begin{aligned} v_y & =u_y+a_y t \\ & =0+10 \times 1 \\ & =10 \mathrm{~m} / \mathrm{s} \end{aligned}$$ Horizontal displacement after $1 \mathrm{~s}$, $$\begin{aligned} s_X & =u_x t+\frac{1}{2} a_X t^2 \\ & =20 \times 1+0=20 \mathrm{~m} \end{aligned}$$ Vertical displacement after $1 \mathrm{~s}$, $$\begin{aligned} s_y & =u_y t+\frac{1}{2} a_y t^2 \\ & =0+\frac{1}{2} \times 10 \times(1)^2=5 \mathrm{~m} \end{aligned}$$ Resultant velocity after $1 \mathrm{~s}$, $$\begin{aligned} v= & \sqrt{v_x^2+v_y^2} \\ = & \sqrt{(20)^2+(10)^2}=\sqrt{400+100}=\sqrt{500} \\ & =22.36=22.4 \mathrm{~m} / \mathrm{s}, \end{aligned}$$ So, the correct option is (c).