An arrow is shot into the air at a velocity of 25 m/s and an angle of 32 degrees above horizontal. It lands at the same level that it was shot from. How long is the arrow in air? How far does the arrow travel?
L=vₒ²•sin2α/g,
t= 2vₒ•sinα/g
WHAT EVEN
To find the time the arrow is in the air, we can use the vertical component of its velocity. The initial velocity in the vertical direction is given by V_y = V * sin(theta), where V is the initial velocity (25 m/s) and theta is the angle of elevation (32 degrees).
V_y = 25 m/s * sin(32 degrees)
V_y = 25 m/s * 0.52992 (rounded to five decimal places)
V_y ≈ 13.248 m/s
Since the arrow landed at the same level it was shot from, we know the vertical displacement is zero. We can use the kinematic equation:
Δy = V_y * t + (1/2) * g * t^2
Where Δy is the vertical displacement, V_y is the vertical component of the initial velocity, t is the time in the air, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
0 = 13.248 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Since we want the time when the arrow is in the air, we solve this quadratic equation to find the time t. Let's rewrite the equation in standard form:
(1/2) * 9.8 m/s^2 * t^2 + 13.248 m/s * t = 0
Solving this quadratic equation, we get two possible solutions for t:
t = 0 (no time spent in the air)
t = -26.99 s (impossible, as time cannot be negative)
Since time cannot be negative, the arrow spends 0 seconds in the air. This means it lands immediately after being shot.
To find the horizontal distance traveled by the arrow, we can use the horizontal component of its velocity. The initial velocity in the horizontal direction is given by V_x = V * cos(theta), where V is the initial velocity (25 m/s) and theta is the angle of elevation (32 degrees).
V_x = 25 m/s * cos(32 degrees)
V_x = 25 m/s * 0.84805 (rounded to five decimal places)
V_x ≈ 21.201 m/s
Since the time in the air is 0 seconds, the horizontal displacement is simply the product of the horizontal velocity and the time:
Δx = V_x * t
Δx = 21.201 m/s * 0 s
Δx = 0 m
Therefore, the arrow does not travel any horizontal distance and lands at the same point from where it was shot.