A projectile is shot from the ground at an angle of #pi/6 # and a speed of #3 /2 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer
Jan 14, 2016

#s approx 0.099 m#

Explanation:

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Step 1: Maximum height will be reached when #sin theta# component of projectile's velocity becomes 0 (zero)

Use the formula
#v=u+g t#
Given #v=0, u= 3/2 . sin (pi / 6) #; as #theta=pi/6#
Take #g= 9.81 m/s^2#
Since gravity is acting against the velocity so it is deceleration and #-# sign is used in front of g
#t= (3/2 . sin (pi / 6))/9.81 s#

Step 2:Distance traveled horizontally in time t is due to #cos theta# component of the initial velocity of the projectile. Assuming zero air friction, and using the equation

#s=ut#
#s= 3/2 . cos (pi / 6) times (3/2 . sin (pi / 6))/9.81 m#