A boy is playing a ball in a garden surrounded by a wall 2.5 m high and kicks the ball vertically up from a height of 0.4 m with a speed of 14 m/s. For how long is the ball above

the height of the wall.

2.54 seconds

Step-by-step explanation:

We can use the following equation to model the vertical position of the ball:

S = So + Vo*t + a*t^2/2

Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.

Then, using S = 2.5, So = 0.4, Vo = 14 and a = - 9.8 m/s2, we have that:

2.5 = 0.4 + 14*t - 4.9t^2

4.9t^2 - 14t + 2.1 = 0

Solving this quadratic equation, we have that t1 = 2.6983 s and t2 = 0.1588 s.

Between these times, the ball will be higher than 2.5 m, so the amount of time the ball will be higher than 2.5 m is:

t1 - t2 = 2.6983 - 0.1588 = 2.54 seconds