Velocity of a stone makes an angle 53° with the ground at an instant. After 2 s, it reaches the maximum height. What will be the velocity of the stone at highest point?

From the concept of projectile motion, the verticle speed at time t is given by:

$V_y=v_y-gt$

where $v_y$ is the initial speed and $V_y$ is the final speed.

At the maximum height its $V_y=0\text{, }$ 

hence at maximum height, we can get the below equation:

$\begin{array}{l}\Rightarrow 0=v_y-gt\\ \Rightarrow v_y=gt\\ \Rightarrow v_y=g\times 2=20\mathrm{m}/\mathrm{s}\end{array}$

also, from the question, $\theta ={53}^{\circ },$ and the situation is shown in the given diagram:

hence,

$\begin{array}{l}\Rightarrow \tan \theta =\frac{v_y}{v_x}\\ \Rightarrow \tan {53}^o=\frac{20}{v_x}\\ \Rightarrow v_x=15{\mathrm{ms}}^{-1}\end{array}$

Hence the required velocity at the maximum height

$v=v_x=15\mathrm{m}/\mathrm{s}$