If a projectile is shot at an angle of #pi/2# and at a velocity of #64 m/s#, when will it reach its maximum height??

1 Answer
Mar 10, 2018

Answer:

It reaches its highest point #6.5 s# after launch.

Explanation:

#pi/2# is straight up so we do not need to calculate a vertical component of the initial velocity. We can use the kinematic formula

#v = u + a*t#

where #v = 0# because it stops at maximum height,
#u = 64 m/s#,
and #a = -9.8 m/s^2# which is the acceleration due to gravity, g. The value of a gets a minus sign because it is in the opposite direction of the initial velocity.

#v = u + a*t#

#0 = 64 m/s - 9.8 m/s^2*t#

Solving for t,

#t = (64 cancel(m)/s)/(9.8 cancel(m)/s^2)#
(hmmm - I struggled with the formatter here. I wanted to strike out the power of 2 on the s, leaving just s. Would not work for me.)

# t = 64 /9.8 s = 6.53 s ~= 6.5 s#

I hope this helps,
Steve