Question

A missile is launched from the ground making an angle of 45° with the horizontal to hit a target at a horizontal distance of 300 km. If it is required to hit a target at a horizontal distance of 675 km launched at same angle with horizontal, find the percentage change in its velocity.

Please do not give any link or hint for this question and provide the solution with steps.

Answer 1

the ans will be 7892.35 m/sec

Answer 2

Case 1:

Range = uCosθ × 2uSinθ/g

=> 300 = uCos45° × 2uSin45°/g

=> 300 = u/√2 × 2u/√2/10

=> 300 = u/√2 ×√2/10

=> 300 = u^2/10

=> 3000 = u^2

=> u = √3000 = 54 m/s

Case 2:

Range = uCosθ × 2uSinθ/g

=> 675= uCos45° × 2uSin45°/g

=> 675= u/√2 × 2u/√2/10

=> 675= u/√2 ×√2/10

=> 675= u^2/10

=> 6750 = u^2

=> u = √6750 = 82 m/s

Difference in velocity= 82 - 54 = 27 m/s

Percentage of change in V = $\frac{26}{54}$ × 100

= 50%