A projectile is shot at a velocity of #9 m/s# and an angle of #pi/12 #. What is the projectile's peak height?

Question

1268 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-23T20:53:03

#0.27679m#

Data:-

Initial Velocity#=# Muzzle Velocity#=v_0=9m/s#

Angle of throwing #=theta=pi/12#

Acceleration due to gravity#=g=9.8m/s^2#

Height#=H=??#

Sol:-

We know that:

#H=(v_0^2sin^2theta)/(2g)#

#implies H=(9^2sin^2(pi/12))/(2*9.8)=(81(0.2588)^2)/19.6=(81*0.066978)/19.6=5.4252/19.6=0.27679#

#implies H=0.27679m#

Hence, the height of the projectile is #0.27679m#