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ana9340
22.06.2020 •
Mathematics
A boy is playing ball in a garden surrounded by a wall 2.5 m high kicks the ball vertically up from a height of 0.4 m with a spee d of 14 m/s.For how long is the ball above the height of the wall .
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Ответ:
2.5 sec
Step-by-step 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)
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
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![t^{2}](/tpl/images/0691/7263/7de90.png)
7.49 = (0 x t) +
(9.81 x
)
7.49 = 0 + 4.9![t^{2}](/tpl/images/0691/7263/7de90.png)
t =
= 1.23 sec
Total time spent above wall = 1.27 s + 1.23 s = 2.5 sec
Ответ: