If a projectile is shot at an angle of #(3pi)/8# and at a velocity of #2m/s#, when will it reach its maximum height?

Question

1326 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-26T06:13:45

The time is #=0.19s#

The initial speed is #u=2ms^-1#

The angle is #theta=3/8pirad#

The acceleration due to gravity is #g=9.8ms^-2#

The equation of the trajectory of the projectile is

#y=xtantheta-(gx^2)/(2u^2cos^2theta)#..............#(1)#

At the maximum height

#dy/dx=0#

#=>#, #dy/dx=tantheta-(2gx)/(2u^2costheta)#

#=tantheta-(gx)/(u^2cos^2theta)#

Therefore,

#tantheta-(gx)/(u^2cos^2theta)=0#

#x=(u^2tanthetacos^2theta)/(g)=u^2(sinthetacostheta)/g#

The constant horizontal velocity is

#v_x=ucostheta#

Therefore,

The time to reach the greatest height is

#t=x/v_x=(u^2(sinthetacostheta)/g)*1/(ucostheta)#

#=(usintheta)/g#

#=2*sin(3/8pi)/9.8#

#=0.19s#