physics kinematics 2d

Question

A projectile was launced horiontally with a velocity of 388 m/s , 2.89 m above the ground. How long did it take the projectile to hit the ground?

Answer 1

Solution:y

What we know: X = traveled Horizontal distance = ? , V_ix = initial velocity in the horizontal direction = 388 m/s, a_x = Horizontal acceleration = 0 m/s ,Y = traveled vertical distance = 2.89 m , Viy = initial vertical Velocity = 0 m/s, ay = gravitational acceleration = 9.81 m/s²

We need Time of travel = t = ?

Starting from the Vertical equation of motion Y = V_iy t + 0.5 ( a_y) t², - 2.89 m = (0m/s)(t) + 0.5 ( -9.81m/s²) t²

-2.89 m = -4.9 t² , t² = 0.59 s² , t = 0.77 s

Answer 2

If you take coordinate system xoy with oridjen onthe ground then equetion of vertical motion y = y₀ - 1/2gt²..

At final moment y = 0. We get equation for t: y₀ - 1/2gt² = 0. Here y₀ = 2.89 m and g = 9.8 m/sec².

t = (2y₀/g)^1/2 = (2·2.89m/9.8m/sec²)^1/2 = .765 sec