Question

14. For a projectile, the ratio of maximum height reached
square of flight time is (g= 10 ms )
(1) 5:4
(2) 5:2
(3) 5:1
(4) 10:1​

Answer 1

Explanation:

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Answer 2

Answer:

(1) 5:4

Explanation:

We know that Maximum height (Hmax) is given as;

Hmax = (U²Sin²∅)/2g

And the flight time is given as;

T = 2USin∅/g

Now the ratio of Hmax and T² will be;

Hmax ÷  T² =  (U²Sin²∅)/2g  ÷ (2USin∅/g)²

⇒ (U²Sin²∅)/2g  ÷ (4U²Sin²∅)/g²

⇒ (U²Sin²∅)/2g  × g²/(4U²Sin²∅)

Simplifying further gives

⇒ g/8 {where g=10m/s}

: .  Hmax ÷  T² = 10/8 = 5/4

So therefore the ratio of maximum height reached

square of flight time will be 5:4