I will help you but not do all for you. $\displaystyle a_y(t)=-9.8 m/s^2$.
$\displaystyle v_y(t)=-9.8t+v_{0y}=-9.8t+24\sin 19°$ so
$\displaystyle y(t)=-4.9t^2+24\sin 19°t$.
While for the x-axis, $\displaystyle a_x(t)=0$.
$\displaystyle v_x(t)=v_{0x}=24\cos 19°$ so $\displaystyle x(t)=24\cos 19°t$.
Maximum height is when $\displaystyle v_{0y}=0$, just solve it.
Hang-time is when you solve for t when $\displaystyle y(t)=0$.
And maximum range is when $\displaystyle y(t)=0$, solve for $\displaystyle t$ and substitute this $\displaystyle t$ into $\displaystyle x(t)$.
And don't forget the units.
__________________ **Isaac** If the problem is too hard just let the Universe solve it. |