A ball is projected horizontally with a velocity of 5 ~ms ⁻¹ from the top of a building 19.6 ~m high. How long will the ball take to hit the ground?

Step-by-step solution

The motion of the ball in the vertical direction is governed by the equation of motion under constant acceleration due to gravity g = 9.8 ~ms ⁻² . The vertical displacement h is given by h = u_y t + 1 2 g t^2 , where u_y is the initial vertical velocity. Since the ball is projected horizontally, the initial vertical velocity u_y = 0 . Substituting the given values h = 19.6 ~m and g = 9.8 ~ms ⁻² into the equation: 19.6 = 0 t + 1 2 9.8 t^2 19.6 = 4.9 t^2 t^2 = 19.6 4.9 t^2 = 4 t = 2 ~s Answer: 2 ~s