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.
Explanation:
At maximum height, the vertical component of a ball's speed is 0. So the net speed is equal to whatever the horizontal component of the ball is. If the ball is launched straight up, the speed at the highest point is simply 0.
Forces equal the flag doesn't move either direction
Unequal forces the flag moves in the direction of the greater magnitude force
Answer:
Zero or +2
Explanation:
The noble gases already have a avplete outermost shell. They are the least reactive elements of earth?
Their normal oxidation number is zero but some have been shown to be reactive.
Answer:
Mass remains constant but weight reduces
Explanation:
Mass is the amount of matter in an object so whether on moon or any other planet, it does not change despite the changes in acceleration.
Weight is a product of mass and acceleration due to gravity, expressed as W=mg where m is the mass, W is weight and g is acceleration. From the above formula, it is evident that when you decrease g, then W also decreases while m is constant. Similarly, when m is constant and g is increased then W also increases.
Therefore, for this case, since g decreases, the weight decreases but mass remains constant.
