An arrow is shot at 29.0° above the horizontal. Its initial speed is 54 m/s and it hits the target.
(a) What is the maximum height the arrow will attain?
(b) The target is at the height from which the arrow was shot. How far away is it?
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4.77 IS DA ANSA
To answer these questions, we can use the equations of projectile motion. In this case, we'll assume the arrow is fired on a level ground with no air resistance.
(a) To find the maximum height the arrow will attain, we need to determine the vertical component of its initial velocity. We can use the following equation:
v_y = v * sin(theta)
where v_y is the vertical component of the velocity, v is the initial speed of the arrow, and theta is the launch angle above the horizontal.
Applying the values given in the problem:
v = 54 m/s
theta = 29 degrees
First, convert the angle to radians:
theta_radians = theta * (pi/180) = 29 * (pi/180) = 0.506 rad
Now, calculate the vertical component of the velocity:
v_y = v * sin(theta_radians) = 54 * sin(0.506) ≈ 28.318 m/s
The maximum height can be found using the equation for vertical displacement (h_max):
h_max = (v_y^2) / (2 * g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values:
h_max = (28.318^2) / (2 * 9.8) ≈ 42.781 m
Therefore, the maximum height the arrow will attain is approximately 42.781 meters.
(b) To find the distance to the target, we need to calculate the horizontal component of the velocity and the time of flight.
The horizontal component of the velocity (v_x) remains constant throughout the motion. We can find it using the following equation:
v_x = v * cos(theta)
Plugging in the values:
v_x = 54 * cos(0.506) ≈ 43.794 m/s
Now, to find the time of flight (t), we can use the vertical equation of motion:
h = v_y * t - (1/2) * g * t^2
Since we know the initial vertical velocity (v_y), the acceleration due to gravity (g), and the maximum height (h_max), we can rearrange the equation to solve for the time of flight:
h_max = v_y * t - (1/2) * g * t^2
Rearranging this equation:
(1/2) * g * t^2 - v_y * t + h_max = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± sqrt(b^2 - 4*a*c)) / (2*a)
where a = (1/2) * g, b = -v_y, and c = h_max.
Plugging in the values:
a = (1/2) * 9.8 = 4.9 m/s^2
b = -28.318 m/s
c = 42.781 m
t = (-(-28.318) ± sqrt((-28.318)^2 - 4*(4.9)*(42.781))) / (2*(4.9))
Calculating this equation will give you two possible values for t: one representing the time at which the arrow reaches the maximum height and another representing the total time of flight. We will take the greater value as the total time of flight.
Using the positive sign in the quadratic formula:
t = (28.318 ± sqrt((28.318)^2 - 4*(4.9)*(42.781))) / (2*(4.9))
Simplifying this will give us the total time of flight, t.
Finally, to find the distance to the target (d), we can use the equation:
d = v_x * t
Plugging in the previously calculated values for v_x and t, we can find the distance.
Please note that the final solution for t may provide two possible values: one representing the time at which the arrow reaches the maximum height and another representing the total time of flight. We will use the greater value as the total time of flight to calculate the distance.