Question

a frog sees an insect at a horizontal distance of 1meter from it .the frog can jump with a maximum speed of 4m/s in any direction .find the mimimum distance that the insect has to keep the frog so that it can escape from the frog

Answer 1

I do not know if you have derived the maximum horizontal distance for a projectile is when fired at 45 degrees up from horizontal. For now I assume you know
then
range = S cos 45 t where S is speed at launch and t is time in the air = 1 meter
now for time in the air
Vi = S sin 45 initial speed up
h = 0 + Vi t - 4.9 t^2
hits ground when h = 0
Vi t = 4.9 t^2
t = 0 (at takeoff) or t = Vi/4.9 = S sin 45 /4.9
note : useful to remember t = 2 Vi/g and h =Vi^2/2g
so
range = S cos 45 * S sin 45 /4.9 = S^2/9.8

Answer 2

now assuming not 45, call angle up A
u = S cos A
Vi = S sin A
range = R = S cos A * t
t = 2 Vi/g = 2 S sin A /g
so
R = [2/g S^2] sin A cos A = (S^2/g) sin 2A
but sin 2A is never bigger than 1
and that is when 2A = 90 degrees
so for max range angle A = 45 degrees (like we assumed)

Answer 3

Your answer is not having a clarity
Please do write the reasons and basic equations so that everyone can understand