Question

The highest barrier that a projectile can clear is $20m$, when the projectile is launched at an angle of $40.0$ degrees above the horizontal. What is the projectile's launch speed? (in m/s)

• A. $26$
  
• B. $25.5$
  
• C. $30.9$
  
• D. $40.9$
  

Answer 1

Answer: (C)

$30.9$

Given : $H=20m$ $\theta ={40}^o$
Maximum height achieved $H=\frac{u^2{\sin }^2\theta }{2g}$
where $sin40=0.64$
$\therefore$ $20=\frac{u^2\times (0.64{)}^2}{2\times 9.8}$
$⟹u=30.9m/s$