Explanation:
At the maximum height, the ball's velocity is 0.
v² = v₀² + 2a(x - x₀)
(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)
x = 7.72 m
The ball reaches a maximum height of 7.72 m.
The times where the ball passes through half that height is:
x = x₀ + v₀ t + ½ at²
(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²
3.86 = 12.3 t - 4.9 t²
4.9 t² - 12.3 t + 3.86 = 0
Using quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8
t = 0.368, 2.14
The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.
Answer:
a. normal
Explanation:
In the field of physics the normal is a line drawn at a right angle to a barrier. In other words the normal line is the line that is drawn perpendicular (right angle, 90 degrees) to the reflective surface of a mirror, or the particular boundary in which refraction occurs at the point of incidence of a light ray. This can be seen in the picture attached below.
Answer:
after 6 second it will stop
he travel 36 m to stop
Explanation:
given data
speed = 12 m/s
distance = 100 m
decelerates rate = 2.00 m/s²
so acceleration a = - 2.00 m/s²
to find out
how long does it take to stop and how far does he travel
solution
we will apply here first equation of motion that is
v = u + at ......1
here u is speed 12 and v is 0 because we stop finally
put here all value in equation 1
0 = 12 + (-2) t
t = 6 s
so after 6 second it will stop
and
for distance we apply equation of motion
v²-u² = 2×a×s ..........2
here v is 0 u is 12 and a is -2 and find distance s
put all value in equation 2
0-12² = 2×(-2)×s
s = 36 m
so he travel 36 m to stop
