Question

A projectile is launched with initial velocity of (20i cap- 30jcap) m/s (taking horizontal direction as x-axis and vertically upward as positive y-axis). With what velocity it will reach the level of projection?​

Answer 1

Answer:

9.6 m

this is the correct answer of this question

Answer 2

given initial velocity of a projectile, find the velocity when it reaches the level of projection

Explanation:

1. given initial velocity $(20i-30j)m/s$ then the initial angle of projectile is given by, $sin\theta= \frac{v_y}{\sqrt{v_x^2+v_y^2}} = \frac{-3}{\sqrt{13} } \\cos\theta=\frac{v_x}{\sqrt{v_x^2+v_y^2}}=\frac{2}{\sqrt{13} }$                
2. hence, the total time of flight in projectile is given by,                                              $T=|\frac{2v_o}{g}(sin\theta)| \ \ \ \ \ \ \ \ (v_o=\sqrt{v_x^2+v_y^2}=10\sqrt{13}m/s)$                                                               $T=|\frac{2(10\sqrt{13} )}{10}(\frac{-3}{\sqrt{13} } )|\ \ \ \ \ \ (taking\ g=10m/s^2) \\T=6\ s$  
3. now , velocity when it reaches the level of projection is given by ,                                  $v_x=v_ocos\theta, \ \ \ \ \ \ \ \ v_y=v_osin\theta-gT$                                                                                      $v_x=10\sqrt{13} (\frac{2}{\sqrt{13} }) = 20\ m/s\\v_y=10\sqrt{13} (\frac{-3}{\sqrt{13} } )-10(6)= -90\ m/s$        
4. hence, the velocity at the level of projection is $(20i-90j)\ m/s$