Question

find the minimum velocity for which horizontal range of projectile is 39.2m​

Answer 1

Answer:

The range R of a projectile launched at an angle θ with a velocity V is:

R = V^2 Sin2θ / g

The height H reached is:

H = V^2 Sin^2(θ) / 2g

The maximum range occurs when θ = π/4 = 45°, because then 2θ = 90° so Sin(2θ) = 1 (which is the maximum value for Sin of any angle).

In this case, we want R to be 39.2m, so

39.2 = (V^2)(1)/9.81

V = √(39.2 x 9.81) = 19.61m/s

The minimum launch velocity is 19.61m/s