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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:
The magnitude of the acceleration is 
The direction is
i.e the negative direction of the z-axis
Explanation:
From the question we are that
The mass of the particle 
The charge on the particle is 
The velocity is 
The the magnetic field is 
The charge experienced a force which is mathematically represented as

Substituting value



Note :

Now force is also mathematically represented as

Making a the subject

Substituting values



The units which would need to be converted before being used for a calculation are cm and 0 ks.
<h3>
What is Unit?</h3>
This is referred to a standard which is used to make comparisons in the aspect of measurement.
The units cm and 0 ks aren't in their standard form which is m and s respectively.
Read more about Unit here brainly.com/question/4895463
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Answer:
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