Question

The maximum value of the horizontal range for a projectile projected with a velocity of 98 m/sec is

Answer 1

Answer:

980 m

Explanation:

Given-----> Initial velocity of projectile is 98 m/sec

To find-----> Maximum value of the horizontal range.

Solution-----> We know that, formula for horizontal range is

R = ( u² Sin2θ ) / g

For maximum range for given initial velocity , value of Sin2θ , must be maximum and we know that, maximum value of Sin2θ is 1 .

So, putting Sin2θ = 1 , in formula of range

Rₘₐₓ = u² ( 1 ) / g

=> Rₘₐₓ = u² / g

Putting u = 98 m/sec in it , we get,

=> Rₘₐₓ = ( 98 )² / 9.8

= 98 × 98 / 9.8

= 98 × 980 / 98

=> Rₘₐₓ = 980 m

Additional information----->

1) Time of flight = 2u Sinθ / g

2) Maximum height = u² Sin²θ / 2g

Answer 2

Answer:

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