A hawk flying at 25 m/s at an altitude of 150 m accidentally drops its prey. The parabolic trajectory of the falling prey is described parametrically by
x=25t,y=150−4.9t^2
until it hits the ground. The variable x represents the horizontal distance traveled by the prey and the variable y represents its height above the ground. Calculate the distance traveled by the prey from the time it is dropped until it hits the ground. Express your answer correct to the nearest tenth of a meter.
when it hits the ground , y = 0
0 = 150 -4.9t^2
t^2 = 150/4.9
t = √(150/4.9) = appr 5.53
then x = 25(5.53) = appr 138.3 m
To find the distance traveled by the prey from the time it is dropped until it hits the ground, we can first find the time it takes for the prey to hit the ground.
When the prey hits the ground, its height y is equal to 0. Therefore, we can set y = 0 in the equation y = 150 - 4.9t^2 and solve for t:
0 = 150 - 4.9t^2
Rearranging the equation, we get:
4.9t^2 = 150
Dividing both sides by 4.9, we get:
t^2 = 30.6
Taking the square root of both sides, we get:
t = sqrt(30.6) approximately equal to 5.53 (rounded to two decimal places)
Now that we have the time it takes for the prey to hit the ground, we can substitute this value into the equation for x:
x = 25t
x = 25 * 5.53
x = 138.25 (rounded to two decimal places)
Therefore, the distance traveled by the prey from the time it is dropped until it hits the ground is approximately 138.25 meters.