# If a projectile is thrown with a velocity of 19.6 m/s making angle of 30 with x axis then the time taken to reach the highest point is ?

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Al E. Share
Feb 8, 2018

Consider that in projectile motion, $v = 0$ at the height of the parabolic trajectory.

Moreover, consider,

$a = - g$

${v}_{0} = \frac{19.6 m}{s}$

v_(0x) = (19.6m)/s*cos(30°) = (17.0m)/s, and,

v_(0y) = (19.6m)/s*sin(30°) = (9.8m)/s

With these data, we can use this equation regarding the $y$-direction,

$v = {v}_{0 y} + a t$

Hence,

$0 = \frac{9.8 m}{s} - \frac{9.8 m}{s} ^ 2 \cdot t$
$\therefore t \approx 1.0 s$

is the time it takes for the projectile to reach its peak.

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