The horizontal speed of the object 1.0 seconds later is 1) 5.0 m/s.
Explanation:
The motion of an object thrown horizontally off a cliff is a projectile motion, which follows a parabolic path that consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
This means that the horizontal speed of an object in projectile motion does not change, and remains constant during the whole motion.
Since in this case the object has been launched with a horizontal speed of
v = 5.0 m/s
this means that this speed will remain constant during the motion, so its horizontal speed 1.0 s later is also 5.0 m/s.
Learn more about projectile motion:
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Answer:
Explanation:
7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s
7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m
or
h = (45sin14.5)² / (2(9.81)) = 6.47 m
which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.
Answer: The correct answer is graph A.
Explanation:
See Khan Academy.
Previous results tell us the speed (v) is given in terms of the coefficient of friction (k) and the radius of the curve (r) as
v = √(kgr)
v = √(0.20·9.8 m/s²·50 m)
= 7√2 m/s ≈ 9.90 m/s
