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.
I think it’s D sorry If I’m wrong
The black means that it is a great emitter/absorber of the electromagnetic spectrum. The electromagnetic radiation is reflected of the white and absorbed nurture black meaning that the temperature of the black tarmac increases to that greater the the white
Answer:
it look the same just to tell you
Answer:
54 km/hr
Explanation:
m/s to km/hr => 18/5
15 m/s to km/hr => 15 x 18/5 =>3 x 18 => 54km/hr