A distress flare is shot straight up from a ship’s bridge 75 ft above the water with an initial
velocity of 76 ft/sec.
a) Write a set of parametric equations for the path of the flare.
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What window gives you the best picture of the problem situation?
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b) What is the height of the flare after 2 seconds? __________________________
(a)
x(t) = 0 (since it went straight up)
y(t) = 75 + 76t - 16t^2
(b) Just plug in t=2
a) To write a set of parametric equations for the path of the flare, we can consider the motion in the vertical and horizontal directions separately.
Let's assume the flare starts at time t=0 and its initial position is (0, 75) where 75 is the height above the water.
In the vertical direction, the acceleration due to gravity can be considered as -32 ft/s^2 (assuming upwards positive and using the standard value for acceleration due to gravity). The initial velocity in the vertical direction is 76 ft/s.
Using the kinematic equation, we can determine the height of the flare as a function of time (t) using the equation:
y = 75 + (76t) + (1/2)(-32)(t^2)
In the horizontal direction, there is no acceleration, so the horizontal velocity remains constant at all times. Let's assume the velocity in the horizontal direction is v ft/s.
Using the definition of velocity as distance divided by time, we can express the horizontal position of the flare (x) in terms of time (t) using the equation:
x = vt
Combining these equations, we can write the set of parametric equations for the path of the flare:
x = vt
y = 75 + (76t) - (16t^2)
b) To find the height of the flare after 2 seconds, we can substitute t=2 into the equation for y:
y = 75 + (76*2) - (16*2^2)
y = 75 + 152 - 64
y = 163 ft
Therefore, the height of the flare after 2 seconds is 163 ft.