A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the
1 answer:
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Answer
Time period T = 1.50 s
time t = 40 s
r = 6.2 m
a)
Angular speed ω = 2π/T
=
= 4.189 rad/s
Angular acceleration α =
=
= 0.105 rad/s²
Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²
b)The maximum speed.
v = 2πr/T
=
= 25.97 m/s
So centripetal acceleration.
a =
=
= 108.781 m/s^2
= 11.1 g
in combination with the gravitation acceleration.
Answer:
They will meet at a distance of 7.57 m
Given:
Initial velocity of policeman in the x- direction,
The distance between the buildings,
The building is lower by a height, h = 2.5 m
Solution:
Now,
When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.
Thus
From the second eqn of motion, we can write:
t = 0.707 s
Now,
When the policeman was chasing across:
The distance they will meet at:
9.57 - 2.0 = 7.57 m