Answer:
V₀y = 0 m/s
t = 2.47 s
V₀ₓ = 61.86 m/s
Vₓ = 61.86 m/s
Explanation:
Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL
<u>V₀y = 0 m/s</u>
For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:
Y = V₀y*t + (0.5)gt²
where,
Y = Height = 30 m
g = 9.8 m/s²
t = time to hit the ground = ?
Therefore,
30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²
t² = 30 m/4.9 m/s²
t = √6.122 s²
<u>t = 2.47 s</u>
For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,
V₀ₓ = Xt
where,
V₀ₓ = Initial vertical Velocity = ?
X = Horizontal Distance = 25 m
Therefore,
V₀ₓ = (25 m)(2.47 s)
<u>V₀ₓ = 61.86 m/s</u>
<u></u>
Due, to uniform motion in horizontal direction:
Final Vertical Velocity = Vₓ = V₀ₓ
Vₓ = 61.86 m/s
Answer:
v(t) = 27 units
Explanation:
The function s(t) represents the position of an object at time t moving along a line such that,
and
We need to find the average velocity of the object over the interval of time [2,6]. The velocity of the object is equal to the total distance divided by time. It is given by :
v(t) = 27 units
So, the average velocity of the object is 27 units. Hence, this is the required solution.
Answer:
0.208 N
Explanation:
We are given that
Distance,d=0.41 m
The magnitude of the net electrostatic force experienced by any charge at point 4
Net force,
Where 
The answer: Solanum tuberosum
