Question

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

Answer 1

Answer:

the time taken to reach the maximum height is 5 s.

Step-by-step explanation:

The final velocity of the body thrown = 50 m/s

The initial velocity of the body = 0 m/s

The time taken to reach maximum height = t s

The maximum height reached by the body = s m

The acceleration due to gravity = 10 m/s²

Using equations of motion :

v² - u² = 2as

⇒ 50² - 0² = 2×10×s

⇒ 2500 = 20s

⇒ s = 2500/20

⇒ s = 250/2

⇒ s = 125 m

Hence, the maximum height reached by the body is 125 m.

s = ut + ½at²

⇒ 125 = 0×t + ½×10×t²

⇒ 125 = ½×10×t²

⇒ 125 = 5t²

⇒ t² = 125/5

⇒ t² = 25

⇒ t = √25

⇒ t = 5 s

∴ the time taken to reach the maximum height is 5 s.

Answer 2

Answer:

125m

Step-by-step explanation:

FOR HEIGHT:-

v^2 = u^2 + 2as

0^2 = 50^2 + 2(10)(s)

0 = 2500 + 20S

2500 = 20s

2500/20 = s

125 = s

MAXIMUM HEIGHT IS 125M

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FOR TIME:-

V=U+AT

0 = 50 + 10T

50 = 10T

50/10 =T

5 = T

TIME TAKEN IS 5 SECOND

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