a swimmer dives off a pier at an angle of 25 degrees from the horizontal with an initial velocity of 3.2 m/s. How far from the pier will the diver hit the water. Assume the level of water is the same as the pier.
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To find how far from the pier the diver will hit the water, we can break down the initial velocity into horizontal and vertical components.
Given:
Initial velocity (v) = 3.2 m/s
Angle of dive (θ) = 25 degrees
We can find the horizontal component of the velocity, v_x, using the following equation:
v_x = v * cos(θ)
We can find the vertical component of the velocity, v_y, using the following equation:
v_y = v * sin(θ)
Now, using the kinematic equation:
y = y_0 + v_y * t + (1/2) * a_y * t^2
In this case, the vertical displacement (y) is equal to zero since the water level is the same as the pier. Thus, we can find the time (t) it takes for the diver to hit the water by rearranging the equation as follows:
0 = v_y * t + (1/2) * a_y * t^2
Since the acceleration in the vertical direction is due to gravity, we can substitute a_y with -9.8 m/s^2, representing the downward acceleration due to gravity. Rearranging the equation to solve for t:
(1/2) * -9.8 * t^2 = -v_y * t
Substituting in v_y = v * sin(θ):
(1/2) * -9.8 * t^2 = -(v * sin(θ)) * t
Now, using the horizontal motion equation:
x = x_0 + v_x * t + (1/2) * a_x * t^2
For simplicity, let's assume the initial position (x_0) is zero, and for this horizontal motion, the acceleration (a_x) is also zero since there is no horizontal force acting on the diver.
Simplifying the equation:
x = v_x * t
Substituting the known values:
x = (v * cos(θ)) * t
Now, we can substitute t from the vertical motion equation into the horizontal motion equation:
x = (v * cos(θ)) * [-(v * sin(θ)) * t / (1/2) * -9.8]
Simplifying further:
x = (v^2 * cos(θ) * sin(θ)) / 9.8
Substituting the given values:
x = (3.2^2 * cos(25°) * sin(25°)) / 9.8
Evaluating this expression, we find:
x ≈ 1.23 meters
Therefore, the swimmer will hit the water approximately 1.23 meters away from the pier.