Question #1c727

1 Answer
Feb 8, 2018

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

Moreover, consider,

#a = -g#

#v_0 = (19.6m)/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_(0y) + at#

Hence,

#0 = (9.8m)/s - (9.8m)/s^2 * t#
#therefore t approx 1.0s#

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