The initial speed of the shot is 15.02 m/s.
The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.
Pl refer to the attached diagram.
Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.
Write an expression for R.

Therefore,

In the time t, the net displacement of the shotput is y in the downward direction.
Use the equation of motion,

Substitute the value of t from equation (1).

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

The shot put was thrown with a speed 15.02 m/s.
Answer:
C) one-half as great
Explanation:
We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

In this case, the sphere starts from rest, so
. Replacing the given values and solving for g':

The acceleration due to gravity near Earth's surface is
. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.
90 percent a day to keep things running smoothly
Answer:
mass = 9.7 kg
Explanation:
Weight = Mass x Acceleration due to gravity (g)
16.5 = mass x 1.7
mass =
= 9.7 kg
Answer:
5 kg
Explanation:
Acceleration = 6 m/s^2
Force = 30 N
Force = mass * acceleration
mass = force / acceleration
mass = 30 / 6
mass = 5 kg