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>
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Due, to uniform motion in horizontal direction:
Final Vertical Velocity = Vₓ = V₀ₓ
Vₓ = 61.86 m/s
plane is flying at an altitude of 70 m
now if an object is dropped from it then time taken by object to drop on ground will be given as
here initial speed in vertical direction must be zero as plane is moving horizontal
given that
y = 70 m
a = 9.8 m/s^2
now since the plane is moving horizontally with speed v = 44 m/s
so the horizontal distance moved by the object will be
so the distance moved by the box is 166.3 m
3260÷4=815 which is you average seed
Precisely around 1,800 miles below.
