Answer:
a) y₂ = 49.1 m
, t = 1.02 s
, b) y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² =
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo =
=
- g t
t =
/ g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 =
² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s
Answer:
the forces acting on it must be strong because gravity is pushing the ball down
Explanation:
25 x 10^-5
= 0.00025
25 cm
= 0.00025 km
Answer:
The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.
The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds
Answer:
I believe the answer would be C. point z