25 POINTS PLS ANSWER
1 answer:
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Answer:
0.67 seconds
8.576 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²
Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.
Speed of the horse = 12.8 m/s
Distance = Speed × Time
⇒Distance = 12.8×0.67
⇒Distance = 8.576 m
Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.
Answer:
The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s
Explanation:
The rotational inertia of the merry-go-round = 600 kg·m²
The radius of the merry-go-round = 3.0 m
The mass of the boy = 20 kg
The speed with which the boy approaches the merry-go-round = 5.0 m/s
Where;
= The tangential force
I = The rotational inertia
m = The mass
α = The angular acceleration
r = The radius of the merry-go-round
For the merry go round, we have;
= The rotational inertia of the merry-go-round
= The angular acceleration of the merry-go-round
= The linear velocity of the merry-go-round
t = The time of motion
For the boy, we have;
Where;
= The rotational inertia of the boy
= The angular acceleration of the boy
= The linear velocity of the boy
t = The time of motion
When the boy jumps on the merry-go-round, we have;
Which gives;
From which we have;
The velocity of the merry-go-round, , after the boy hops on the merry-go-round = 1.5 m/s.