Question 1202581
Initial velocity = U , angle ={{{ theta}}}
Here , Angle = 20 degrees, Initial velocity = 35m/s

35*sin 20 degrees   vertical velocity =11.97 m/s

35*cos 20 degrees   horizontal velocity  30.31 m/s


Newtons equation of motion, S= ut+1/2 * gt^2

a)	Time at which it is 0.4 m height

 S=0.4 m,  u=35 sin 20^0
 -4.9t^2 +11.97t-0.4 =0
Solve
	t=2.40s  or   0.034s  Two values since path is parabolic

(b) Find its position parallel to the field at height equal to 0.40 m.
Two horizontal  distance values 
35*cos 20 degrees   horizontal velocity  30.31 m/s

(1)	30.31 *0.034	m			(2)    30.31* *2.40 m


(c) Find the time when it reaches its highest point.
Max height =  u^2/2g  = (11.97/2)^2*9.8 = 7.31 m
d) At what point is the ball at its highest and farthest?
v=u+gt
Final velocity is 0 at maximum height
0 =11.97 +(-9.8) *t
t = 11.97 /9.8=1.22 s

Time of ascent = time of descent

Farthest = 1.22*2= 2.44 s

 *[illustration projectile.png].