If a projectile is shot at an angle of #pi/3# and at a velocity of #17 m/s#, when will it reach its maximum height??

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Answer 1 · 2016-12-04T07:25:34

The answer is #=11.1m#

At the maximum height, the vertical component of the velocity #=0#

So, we use the equation

#v^2=u^2+2as#

#v=0#

#u=u_0sintheta#

#a=-g=-9.8ms^(-2)#

#s=h# is the maximum height

#0=(u_0sintheta)^2-2gh#

#h=(u_0sintheta)^2/(2g)#

#=(17*sin(pi/3))^2/(2*9.8)#

#sin(pi/3)=sqrt3/2#

#h=(17^2*3/4)/19.6=11.1m#