A shell is fired from the ground with an initial velocity of 1.70 x 10^3 m/s at an angle of 55.0 degrees ...?

Question

from the horizontal. Neglecting air resistance, find:


a)The Shell's horizontal range





please save me!! I can only rely on geniuses of this world cause i am so stupid but a simpleton like me would do anything for you later so help me!!

Answer 1

To find the range use the equations of motion:
x(t)=cos*v*t
and
vy(t)=v*sin-g*t

The time to apogee is 1/2 the total flight time.
At apogee vy(t)=0
so
v*sin(55)/g=t
plug 2*t into range
x(t)=2*v^2*cos(55)*sin(55)/g
range=27 x10^3 m

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