Question

If a body projected with a velocity of 49 m/s cannot reach more than 98m vertically, the horiental
distance it cover​

Answer 1

Answer:

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Answer 2

Given :

Velocity of projectile , v = 49 m/s .

Maximum height gained , $H_{max}=98\ m$ .

To Find :

The horizontal distance covered by body .

Solution :

We know , maximum height is given by :

$H_{max}=\dfrac{u^2sin^2\theta}{2g}$

Putting all values we get :

$98=\dfrac{49^2sin^2\theta}{2\times 9.8}\\\\sin^2 \theta=0.8\\\\sin\theta=0.89$

Therefore ,

$sin\ 2\theta=2sin\theta\sqrt{1-\sin^2\theta}\\\\sin\ 2\theta=2\times 0.89\times\ \sqrt{1- 0.89^2}\\\\sin\ 2\theta=0.81$

So , horizontal distance covered is :

$d=\dfrac{u^2sin\ 2\theta}{g}\\\\d=\dfrac{ 49^2\times 0.81}{9.8}\\\\d=198.45\ m$

Therefore , horizontal distance covered by the object is 198.45 m .

Learn More :

Projectile Motion

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