An object is thrown upwards with a speed of 14 m/s. how long does it take to reach a height of 5.0 m above the projection point while descending?
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Ответ:
2.43 s
Explanation:
Using newton's equation of motion.
T = (v-u)/g
Where T = time taken for the object to return to the point of projection , u = initial velocity, v = final velocity, g = acceleration due to gravity.
Given: v =-14 m/s, u = 14 m/s, g = -9.8 m/s²
T = (-14-14)/-9.81
T = 2.85 s
Note: We look for the object's speed at 5.0 m.
using
v² = u²+2gs Equation 1
Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.
Given: u = 14 m/s, g = -9.81 m/s², s = 5.0 m
Substitute into equation 1
v² = 14²+(-9.81×5×2)
v² = 196-98.1
v = √97.9
v = 9.89
We look for the time taken for the velocity to decrease from 14 m/s to 9.89 m/s.
using
v = u+gt
t =(v-u)/g Equation 2
Where t = time taken for the object to decrease it velocity from 14 m/s to 9.89 m/s
Given: v = 9.89 m/s, u =14 m/s g = -9.81 m/s²
t = (14-9.89)/-9.81
t = -4.11/-9.81
t = 0.42 s
Thus,
Time taken to reach 5.0 m above projection point = T-t
=2.85-0.42
2.43 s
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Explanation:
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