Question

An archer shoots an arrow with a velocity of $30m/s$ at an angle of $20$ degrees with respect to the horizontal. An assistant standing on the level ground $30$ m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow. What is the initial speed of the apple and at what time after the arrow is launched should the apple be thrown so that the arrow hits the apple?

• A. $0.1s$
  
• B. $0.01s$
  
• C. $1s$
  
• D. $0.001s$
  

Answer 1

Answer: (B)

$0.01s$

horizontal component of velocity is $30.Cos20m/s$

vertical component of initial velocity is $30.Sin20m/s$

time taken to reach horizontal 30 meter is $\frac{30}{30Cos20}=Sec20=1.0641$

vertical position at that time is given by $y=xTan\theta -\frac{g{x}^2}{2v_0^{2}Co{s}^2\theta }$

so $y=5.3643$

To calculate velocity of apple required

$S=5.3643$, $v=0$, $a=-9.81$, $u=?$

$u=\sqrt{2\times 9.81\times 5.3643}=10.2590$

time taken by apple to reach peak position is $\frac{u}{g}=1.0457$

so time after the arrow is launched should be $1.0641-1.0457=0.0184$

none of the above options are exactly correct. still closest option is option B.