Finding horizontal range from projectile.

[shizupple]

Homework Statement

A shell is fired from the ground with an initial speed of 1670m/s at an initial angle of 51° to the horizontal
(a) Neglecting air resistance, find the shell's horizontal range.
(b) Find the amount of time the shell is in motion.

Homework Equations

dx=vi+.5at^2
dv=.5(vi+vf)t
v=dx/dt
a=dv/dt

The Attempt at a Solution

I used pythagorean theorem to find x and y components:
velocity(x)= 1051m/s
velocity(y)= 1298m/s

I divided the vertical component by acceleration due to gravity to get:
1298/9.81= 132s

I multiplied the time it was in the air by the horizontal component to get:
1051x132= 138732m

I got the wrong answers. I feel like this should work. I think it might be a bit more complicated. Maybe I need to rewrite one of the kinematic equations in terms of sins and cos use substitution to find one of the variables? Some direction on what to do would be great! Thanks.

 

[rl.bhat]

The time t you have found is to reach the maximum height. The time of flight is twice this time. The range is v(x)*2t.

 

[shizupple]

The time t you have found is to reach the maximum height. The time of flight is twice this time. The range is v(x)*2t.

Awesome thanks!

 

[Pheo1986]

you can also find out the range using the equation

range = v^2*sin(2*theta)/gravity