Speed of the projectile at its maximum height is only along horizontal direction
so at highest point

now when he is at half of the maximum height the speed will be in x and y direction both

here it is given that




also we know that

here we know that maximum height is given as



now from above


also we know that angle of projection is


so angle is


Answer:
485520 m
Explanation:
= initial velocity of the projectile = 1360 m/s
= final velocity of the projectile =
=
= 544 m/s
a = acceleraton due to gravity on moon = - 1.6 m/s²
h = Altitude of the projectile
Using the kinematics equation

Inserting the values

h = 485520 m
Answer:
The magnitude of the centripetal force that acts on him
Explanation:
Given that,
Mass = 80.0 kg
Distance = 6.10 m
Speed = 6.80 m/s
We need to calculate the magnitude of the centripetal force that acts on him
Using formula of the centripetal force

Where, F = force
m = mass
v = speed
r = distance
Put the value into the formula


Hence, The magnitude of the centripetal force that acts on him
Answer:
hello the diagram relating to this question is attached below
a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec
b) Force exerted on B2 at P = 39.2 N
Explanation:
Given data:
Co = 150 N-m ,
<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>
at point P ; Co = I* ∝B2'
150 = ( (2*0.5^2) / 3 ) * ∝B2
∴ ∝B2' = 900 rad/sec
hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec
at point 0 ; Co = Inet * ∝B1
150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1
∴ ∝B1 = 180 rad/sec
hence angular acceleration of B1 = 180 rad/sec
<u>b) Determine the force exerted on B2 at P</u>
T2 = mB1g + T1 -------- ( 1 )
where ; T1 = mB2g ( at point p )
= 2 * 9.81 = 19.6 N
back to equation 1
T2 = (2 * 9.8 ) + 19.6 = 39.2 N
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