The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
Because some of the energy is wasted and the amount of energy wasted is based on the efficiency of the machine
Answer:
The velocity of the fish hitting the ground is , v = 45.795 m/s
Explanation:
Given data,
The mass of the fish, m = 5 kg
The height of the bird from the surface, h = 107 m
Using the III equation of motion,
v² = u² + 2gs
<em> v = √(u² + 2gs)</em>
Substituting the values,
v = √(0² + 2 x 9.8 x 107)
= 45.795 m/s
Hence, the velocity of the fish hitting the ground is, v = 45.795 m/s
Answer:
Explanation:
By Snell's law we know at the left surface
now we have
now on the other surface we know that
angle of incidence = 
so again we have
so we have
also we know that
By solving above equation we have
Answer:
The last one " They are equal and act in opposite directions"
Explanation:
Hope u got it right.
