Question

a body is thrown up with a velocity 50m/s calculate the maximum height reached and the time taken to reach the maximum height (a=10m/s2)​

Answer 1

Answer:

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Answer 2

Explanation:

CORRECT QUESTION :-

★ A body is thrown up with a velocity of 50 m/s. Calculate the maximum height reached and the time taken to reach the maximum height { g = -10 m/s². }

GIVEN :-

• Initial velocity , u = 50 m/s.

• final velocity , v = 0.

• acceleration due to gravity , g = -10 m/s².

TO FIND :-

• maximum height , s.

• time taken , t.

SOLUTION :-

★ third equation of motion ★

➳ v² - u² = 2gs.

u = Initial velocity.

v = final velocity.

g = acceleration due to gravity.

s = maximum height.

➳ (0)² - (50)² = 2(-10)s

➳ 0 - 2500 = -20 s

➳ -2500 = -20 s

➳ s = -2500/-20

➳ s = 2500/20

➳ s = 125 m.

★ first equation of motion ★

✭ v = u + gt.

u = Initial velocity.

v = final velocity.

g = acceleration due to gravity.

t = time taken.

➝ 0 = 50 + (-10) t

➝ 0 = 50 - 10 t

➝ 0 - 50 = -10 t

➝ -50 = -10 t

➝ t = -50/-10

➝ t = 50/10

➝ t = 5 sec.

Hence the maximum height is 125 m and and time taken to reach the maximum height is 5 sec.