A ball is thrown vertically upwards with a speed of 24.5m/s. For how long is the ball higher than 29.4m above its initial position? Take acceleration due to gravity to be 9.8m/s^2.
To find how much time the ball spends above 29.4m, we need to find the times at which the ball is at 29.4m above its starting point, and then calculate the difference between these times. Acceleration is constant, so we can use the equations of motion. First write out which quantities we know and what we want to find: u = 24.5 m/s v = a = -9.8m/s2 s = 29.4m t = ? We want an equation involving u, a, s and t. Use s = ut + 0.5at2 -> (substitute known values) -> 29.4 = 24.5t + 0.5(-9.8)t2 -> 29.4 = 24.5t - 4.9t2 -> (rearrange so that we have a quadratic equal to zero) -> 4.9t2 - 24.5t + 29.4 = 0 -> (multiply everything by 10 to get rid of the decimals) -> 49t2 - 245t + 294 = 0 -> (divide everything by 49) -> t2 - 5t + 6 = 0 -> (factorise) -> ( t - 2 )( t - 3 ) = 0 -> t = 2 or t = 3 . Therefore time above 29.4m is given by: time = 3 - 2 = 1 second.
