To solve this problem, we can break it down into horizontal and vertical components.
First, we need to find the initial vertical velocity of the shot put by using the given velocity and angle.
Vertical velocity (Vy) = 35 ft./sec * sin(40°)
Vertical velocity (Vy) ≈ 22.6 ft./sec
Next, we can determine the time it takes for the shot put to reach a height of 40 ft. by using the equation for vertical displacement:
Δy = Voy * t + (1/2) * a * t^2
40 ft = 22.6 ft./sec * t - 16 ft./s^2 * t^2
Rearranging the equation, we get:
16t^2 - 22.6t + 40 = 0
Solving this quadratic equation, we get:
t ≈ 1.51 sec
Now, we can use this time to find the horizontal distance the shot put travels:
Horizontal distance = 35 ft./sec * cos(40°) * 1.51 sec
Horizontal distance ≈ 40 ft
Therefore, it will take approximately 1.51 seconds for the shot put to be a horizontal distance of 40 ft. from the person throwing it.
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. A shot put is thrown upward with a velocity of 35 ft./sec. at a height of 4 ft. and an angle of 40°. How long will it take for the shot put to be a horizontal distance of 40 ft. from the person throwing it? Acceleration due to gravity is 32 ft./s2 . Round your answer to the nearest hundredth. (2 points)
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