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 $
$25cos 53 ^\circ = v cos ⁡ 45 ^\circ $
$\Longrightarrow v=25\times \frac{3}{5}.\sqrt{2}$
$=15\sqrt{2}$
$\Longrightarrow v_{y}=vsin 45 = 15$
Now using $ \, \, \, V_{y}=u_{y}+a_{y}t \, in \, \, y-$ direction,
$-15=25sin 53° $
$-15=25\times \frac{4}{5}-10 \, t$
$\Longrightarrow t=3.5 \, \text{sec}$