Answer:
1.5 u
Explanation:
The range equation is:
R = u² sin(2θ) / g
When u = v, R = 2.25 R.
2.25 R = v² sin(2θ) / g
2.25 u² sin(2θ) / g = v² sin(2θ) / g
2.25 u² = v²
1.5 u = v
Answer:
The minumum speed the pail must have at its highest point if no water is to spill from it
= 2.64 m/s
Explanation:
Working with the forces acting on the water in the pail at any point.
The weight of water is always directed downwards.
The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.
And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.
At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.
Net force = W + (normal force)
But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.
At this point of minimum velocity,
Normal force = 0
Net force = W
Net force = centripetal force = (mv²/r)
W = mg
(mv²/r) = mg
r = 0.710 m
g = 9.8 m/s²
v² = gr = 9.8 × 0.71 = 6.958
v = √(6.958) = 2.64 m/s
Hope this Helps!!!
Answer:
1777.92 m/s
Explanation:
R = Radius of asteroid = 545 km
M = Mass of planet
g = Acceleration due to gravity = 2.9 m/s²
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
Acceleration due to gravity is given by

The expression of escape velocity is given by

The escape speed is 1777.92 m/s
Answer:
1) t=1.743 sec
2)Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)Vf=17.08 m/s
Explanation:
1)From second equation of motion we get
h=Vit+(1/2)gt^2
here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion
14.9=(0)*t+(1/2)(9.8)t^2
t^2=14.9/4.9
t^2=3.040 sec
t=1.743 sec
2) s=Vo*t
Putting values we get
107=Vo*1.743
Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)From third equation of motion we know that
Vf^2-Vi^2=2gh
here Vi=0 m/s,h=14.9 m
Vf^2=Vi^2+2gh=0+2(9.8)(14.9)
Vf^2=292.04
Vf=17.08 m/s