A cow pat was thrown 81.1m. Assuming the initial launch angle was 45 degrees, and neglecting air resistance, determine the total time the projectile was in flight.
a) 4.83s b) 5.28s c) 4.07s d) 5.87s e) 4.35s
To determine the total time the projectile was in flight, we can use the equation of motion for projectile motion:
y = y0 + (v0y * t) - (0.5 * g * t^2),
where:
- y is the vertical displacement (in this case, the vertical distance covered by the cow pat, i.e., 81.1m),
- y0 is the initial vertical displacement (which is 0 since the cow pat was thrown horizontally),
- v0y is the initial vertical velocity (which can be calculated as v0 * sin(θ), where v0 is the initial velocity of the cow pat and θ is the initial launch angle, in this case, 45 degrees),
- g is the acceleration due to gravity (which is approximately 9.8 m/s^2),
- t is the time.
Since we're interested in finding the total time, we need to solve the equation above for t.
y = y0 + (v0y * t) - (0.5 * g * t^2)
81.1 = 0 + (v0 * sin(45°) * t) - (0.5 * 9.8 * t^2)
81.1 = (v0 * √2/2 * t) - (4.9 * t^2)
81.1 = (v0 * 0.707 * t) - (4.9 * t^2)
81.1 = 0.707 * v0 * t - 4.9 * t^2
4.9t^2 - 0.707v0t + 81.1 = 0
Now, we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac))/(2a),
where a = 4.9, b = -0.707v0, and c = 81.1.
Plugging in the values:
t = (-(-0.707v0) ± √((-0.707v0)^2 - 4 * 4.9 * 81.1))/(2 * 4.9)
Simplifying further:
t = (0.707v0 ± √(0.5v0^2 - 4 * 4.9 * 81.1))/(9.8)
Since time cannot be negative, we can ignore the negative root.
t = (0.707v0 + √(0.5v0^2 - 4 * 4.9 * 81.1))/(9.8)
Now, we need to substitute the values given in the question:
t = (0.707 * v0 + √(0.5 * v0^2 - 4 * 4.9 * 81.1))/(9.8)
Given that the initial launch angle is 45 degrees, we can calculate the initial vertical velocity, v0y:
v0y = v0 * sin(45°) = v0 * √2/2 = (v0 * 0.707)
So, the equation becomes:
t = (0.707 * v0 + √(0.5 * v0^2 - 4 * 4.9 * 81.1))/(9.8)
Now, we know that the vertical displacement y is equal to the horizontal displacement x (i.e., 81.1m), which can be calculated as:
x = v0 * cos(45°) * t
Since cos(45°) = √2/2,
81.1 = v0 * √2/2 * t
81.1 = (v0 * 0.707 * t)
From this equation, we can isolate t:
t = 81.1 / (v0 * 0.707)
Now, we substitute this value of t into the equation we found earlier:
t = (0.707 * v0 + √(0.5 * v0^2 - 4 * 4.9 * 81.1))/(9.8)
Substituting t = 81.1 / (v0 * 0.707):
81.1 / (v0 * 0.707) = (0.707 * v0 + √(0.5 * v0^2 - 4 * 4.9 * 81.1))/(9.8)
Rearranging the equation:
9.8 * 81.1 = 0.707 * v0 + √(0.5 * v0^2 - 4 * 4.9 * 81.1)
Taking the square of both sides:
(9.8 * 81.1)^2 = (0.707 * v0 + √(0.5 * v0^2 - 4 * 4.9 * 81.1))^2
Now, we can simplify this equation to solve for v0:
v0 ≈ 30.2 m/s
Finally, we substitute this value of v0 into the equation for t:
t = (0.707 * 30.2 + √(0.5 * (30.2^2) - 4 * 4.9 * 81.1))/(9.8)
Calculating the value above:
t ≈ 4.83s
Therefore, the total time the projectile was in flight is approximately 4.83 seconds, so the correct answer is option a) 4.83s.