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

1 Answer

Answer:

Time #t=(3*sin(pi/8))/9.8=0.1171479895" "#second

Explanation:

For projectiles fired at angle #theta# with initial velocity #v_0#

#y=v_0 sin theta* t+1/2*g*t^2#

We are after the maximum height #y_max#.
We differentiate #y# with respect to #t# then equate #dy/dt=0# because we maximizing height #y#

#y=v_0 sin theta* t+1/2*g*t^2#

#dy/dt=v_0 sin theta *dt/dt+1/2*g*2*t*dt/dt#

#dy/dt=v_0 sin theta *1+1/2*g*2*t*1#

#dy/dt=v_0 sin theta+g*t#

#dy/dt=0#

#v_0 sin theta+g*t=0#

Solving for t

#t=(-v_0 sin theta)/g=(-3*sin 22.5^@)/(-9.8)=0.1171479895" "#second

God bless...I hope the explanation is useful.