An object gets the acceleration 8.0 m/s2 in a vertical direction (y) but 5.0 m/s2 in a horizontal direction (x).
What will be the position of the object at the time t=2.0 s if his initial postiotion was (0,0), initial speed in y-direction 0 and in x-direction 5.0 m/s?
Given answers are: 20m ; 16m
Help please :))
ax = 5 m/s^2
ay = 8 m/s^2
vx = Vi + ax t = 5 + 5 t
vy = 0 + ay t = 8 t
x = Xi + Vi t +(1/2) ax t^2
x = 0 + 5 t + (1/2)5 t^2
y = 0 + 0 t + (1/2)8 t^2
at t = 2
x = 0 + 10 +2.5(4) = 20
y = 0 + 0 + 4 (4) = 16