Answer:
2.5 sec
Explanation:
Height of wall = 2.5 m
initial speed of ball = 14 m/s
height from which ball is kicked = 0.4 m
we calculate the speed of the ball at the height that matches the wall first
height that matches wall = 2.5 - 0.4 = 2.1 m
using = + 2as
where a = acceleration due to gravity = -9.81 m/s^2 (negative in upwards movement)
= + 2(-9.81 x 2.1)
= 196 - 41.202
= 154.8
v = = 12.44 m/s
this is the velocity of the ball at exactly the point where the wall ends.
At the maximum height, the speed of the ball becomes zero
therefore,
u = 12.44 m/s
v = 0 m/s
a = -9.81 m/s^2
t = ?
using V = U + at
0 = 12.44 - 9.81t
-12.44 = -9.81
t = -12.44/-9.81
t = 1.27 s
the maximum height the ball reaches will be gotten with
= + 2as
a = -9.81 m/s^2
0 = + 2(-9.81s)
0 = 196 - 19.62s
s = -196/-19.62 = 9.99 m. This the maximum height reached by the ball.
height from maximum height to height of ball = 9.99 - 2.5 = 7.49 m
we calculate for the time taken for the ball to travel down this height
a = 9.81 m/s^2 (positive in downwards movement)
u = 0
s = 7.49 m
using s = ut + a
7.49 = (0 x t) + (9.81 x )
7.49 = 0 + 4.9
= 7.49/4.9 = 1.53
t = = 1.23 sec
Total time spent above wall = 1.27 s + 1.23 s = 2.5 sec