Answer:
330.5 m
Explanation:
In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .
The maximum height will be calculated as;
where ∝ is the angle of launch = 30°
vi= initial launch velocity = 40 m/s
g= 10 m/s²
h= 40²*sin²40° / 2*10
h={1600*0.4132 }/ 20
h= 661.1/2 = 330.5 m
Answer:a) The linear speed of the rotating merry-go-round is 1.884 m/s.
b) The acceleration of the rotating merry-go-round device is .
Explanation:
a) linear speed of the device:
Distance =
Child seated 1.2 meters from the center, which means that radius ,r = 1.2 meters
Time taken to complete one revolution = 4.0 seconds
The linear speed of the rotating merry-go-round is 1.884 m/s.
b) Acceleration
Linear velocity ,v = 1.884 m/s
The acceleration of the rotating merry-go-round device is .