please help anyone please???

Question

When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of 12.7 m/s, at an angle of 25° above the horizontal.
(a) What is the range of his leap?
___m
(b) For how much time is he in the air?
___s

Answer 1

x(t)=12.7*cos(25)*t

and
vy(t)=12.7*sin(25)-g*t
when vy(t)=0, the greyhound reaches apogee, which is at 50% of total flight time
so
0=12.7*sin(25)-g*t solve for t and double it for the flight time
t=12.7*sin(25)/g

now plug t into x(t) to find the range

x(12.7*sin(25)/g)=
((12.7*cos(25))^2)/g
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