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
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)
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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
To find the minimum distance that the insect needs to keep from the frog in order to escape, we need to determine the maximum distance the frog can cover in the time it takes to reach the insect.
The time it takes for the frog to reach the insect can be found using the formula:
time = distance / speed
Here, the distance is given as 1 meter and the maximum speed of the frog is 4 m/s.
So, the time taken by the frog to reach the insect is:
time = 1 meter / 4 m/s = 0.25 seconds
During this time, the insect needs to move to a safe distance where the frog cannot reach it.
Since the frog can jump in any direction, the maximum distance it can cover in any direction in 0.25 seconds is the maximum speed multiplied by the time:
max distance covered = speed * time = 4 m/s * 0.25 seconds = 1 meter
Therefore, the minimum distance the insect needs to keep from the frog is 1 meter.