Answer:
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)
= + 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
Answer:
Use the slope formula (y2-y1)÷(x2-x1)
Step-by-step explanation:
(-5-(-3))÷2-3
= (-2.5,-1)
30% converted to a decimal is 0.30
310 x 0.3 = 93 students play an instrument
Answer:
Statistical file:
{11, 9, 14, 13, 12, 9, 17, 16}
Minimum: 9
Quartile Q1: 10
Median: 12.5
Quartile Q3: 15
Maximum: 17
Step-by-step explanation:
First part, in quadrilateral WXYZ, WX is parallel to ZY because both sides are equal while the other parallel segments WZ and XY are equal but different side lengths.
The measure of angle Z is 78 because both of the pictures are the same, only rotated. and angle S and Z are acute.
