A projectile is shot at a velocity of # 7 m/s# and an angle of #pi/8 #. What is the projectile's peak height?

Question

1285 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-02T12:28:25

#h_"peak"=0.37" "m#

#"the peak height of a projectile can be calculated using the formula :"#
#h_"peak"=(v_i^2*sin^2 alpha)/(2*g)#

#v_i=7" "m/s#

#alpha=pi/8#

#sin (pi/8)=0.383#

#g=9.81" "m/s^2#

#h_"peak"=(7^2*(0.383)^2)/((2*9.81))#

#h_"peak"=(49*0.147)/(19.62)#

#h_"peak"=0.37" "m#