A projectile is shot at a velocity of # 15 m/s# and an angle of #pi/4 #. What is the projectile's peak height?

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Answer 1 · 2016-01-10T12:11:03

#Height=5.7375m#

Data:-
Muzzle Velocity#=v_0=15m/s#

Angle#=thetapi/4=#

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

Height#=H=??#

Sol:-

We know that:-

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

#implies H=(15^2sin^2(pi/4))/(2*9.8)=((225)(0.707)^2)/19.6=(225*0.4998)/19.6=112.455/19.6=5.7375 m#

#implies H=5.7375m#