Simple Projectile Question

Hi there! Okay the question asked:

An athlete throws a javelin from a height of 1.8 m with a velocity of 21m/s at an upward angle of 45 degrees to the ground. How far is the javelin thrown? Assume it's a particle and there is not air resistance.

So the book's method is to use s= -1.8 u= sin45 x 21= 14.84 a=-9.8 , find v then t by subbing to get t= 3.144.

3.144x14.84=46.7 m.

I used the fact that projectile motion are symmetrical to work out t=1.51 from v=u+at. So I did t=1.51x 2 = 3.02

Now for the remaining part I did v=14.84-9.8t
Sub v into s=(u+v)/2 x t to get quadratic 0= 9.8 t^2 - 29.68t + 3.6
t= 2.9 t=0.1265 If I use t= 0.1265 then add it to t=3.02 , t= 3.1465

I can't tell if what I am doing is right or wrong...I will follow the book's method though just to be safe.

Yes that works, but it's a tad more complicated than it need be. Breaking the problem down:

21Sin45 = 14.85m/s gives the vertical velocity at t=0

We can then work out the height reached above the 1.8m position = 11.243m

Now add 1.8m to that to get the total height fallen = 13.043m

And from that use $t = \sqrt{\frac{2s}{g}}$ to get the time taken to fall to the ground from the apex.

Add that to the 1.51 seconds previously worked out and you get the total flight time of 3.145 seconds and finally the total distance in the horizontal direction.