a stone is thrown with a velocity of 20m/s at an elevation of angle A, given by tan A = 3/4, what horizontal distance does it cover in 2 sec, and what is its height then above the horizontal plane through the point of projection?
solution:
r(right) Ux=VcosA=(4/5)V
r(down) Uy=VsinA=(3/5)V
R(right)distance=speed x time
height=(3/5)x V x t =(3/5) 20 x t =12 x 2 = 24 m (BUT answer is 4 m??)
distance= (4/5)Vt =(4/5)20t =16t =16x2 =32 m (correct approach??)
For the height you are expected to include gravity, so there should be a term $-\frac 12gt^2$. For the horizontal distance you are fine.