Question

John is on top of the building and jack is down. If john throws a ball at an angle of 60 degree and with initial velocity 20 m/s . At what height will the ball reach after 2 s?

Answer 1

consider the motion along the Y-direction

v₀ = initial velocity = 20 sin60 = 17.32 m/s

a = acceleration = - 9.8 m/s²

t = time of travel = 2 sec

Y₀ = initial position at the top of building = 0 m

Y = final position above the top of building

Using the equation

Y = Y₀ + v₀ t + (0.5) a t²

Y = 0 + (17.32) (2) +  (0.5) (- 9.8) (2)²

Y = 15.04 m