Question

A stone is thrown at $25\text{m}/\text{s}$ at $53^o$ above the horizontal. At what time its velocity is at $45^o$ below the horizontal?

• A. $0.5\text{s}$
• B. $4\text{s}$
• C. $3.5\text{s}$
• D. $2.5\text{s}$

Answer 1

Answer: (C)

Horizontal component of velocity, throughout the motion remain constant
using $ucos\theta =vcos\alpha$
$25cos53^{\circ }=vcos45^{\circ }$
$⟹v=25\times \frac{3}{5}.\sqrt{2}$
$=15\sqrt{2}$
$⟹v_y=vsin45=15$
Now using $V_y=u_y+a_{y}tiny-$ direction,
$-15=25sin53°$
$-15=25\times \frac{4}{5}-10t$
$⟹t=3.5\text{sec}$