An object is thrown vertically upward in the sky, it returns to the ground after 6 s. Calculate
(i) the initial velocity of the object
(ii) the maximum height it reaches.
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An object is thrown vertically upward in the sky, it returns to the ground after 6 s. Calculate
(i) the initial velocity of the object
(ii) the maximum height it reaches.
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Answer:
[tex]\bigstar{\bold{Initial\:velocity=29.4\:m/s}}[/tex]
[tex]\bigstar{\bold{Maximum\:height=44.1\:m}}[/tex]
Explanation:
[tex]\Large{\underline{\underline{\sf{Given:}}}}[/tex]
[tex]\Large{\underline{\underline{\sf{To\:Find:}}}}[/tex]
[tex]\Large{\underline{\underline{\sf{Solution:}}}}[/tex]
Initial velocity:
➣ We have to find initial velocity (u) of the object.
➣ By the first equation of motion we know that
v = u + at
➣ Here the time taken for the total journey is 6 s
➣ Hence time taken for upward journey is 6/2 = 3 s
➣ Substituting the datas we get,
0 = u + -9.8 × 3
➣ Here g is taken as negative since the motion of the object is in the direction opposite to that of accleration due to gravity.
➣ Solving it we get,
0 = u - 29.4
u = 29.4 m/s
➣ Hence the initial velocity of the body is 29.4 m/s
[tex]\boxed{\bold{Initial\:velocity=29.4\:m/s}}[/tex]
Maximum height:
➛ By the second equation of motion we know that,
s = ut + 1/2 × a × t²
➛ Substituting the datas,
s = 29.4 × 3 + 1/2 × -9.8 × 3 × 3
s = 88.2 + -88.2/2
s = 88.2 - 44.1
s = 44.1 m
➛ Hence the maximum height reached by the object is 44.1 m
[tex]\boxed{\bold{Maximum\:height=44.1\:m}}[/tex]
[tex]\Large{\underline{\underline{\sf{Notes:}}}}[/tex]
→ The three equations of motion are :