An arrow is shot at an angle 35 ◦ with the horizontal. It has a velocity of 47 m/s. How high will the arrow go? The acceleration of gravity is 9.8 m/s 2 .
Answer in units of m
Vo = 47m/s @ 35o.
Yo = 47*sin35 = 27 m/s = Ver. component
of initial velocity.
Y^2 = Yo^2 + 2g*h.
h = (Y^2-Yo^2)/2g.
h = (0-729)/-19.6 = 37.2 m.
To find the maximum height reached by the arrow, we can use the equations of motion.
First, let's break down the initial velocity into its horizontal and vertical components.
The horizontal component, vx, remains constant throughout the motion and can be found using the velocity and the angle:
vx = v * cos(θ)
vx = 47 * cos(35°)
vx ≈ 38.42 m/s
The vertical component, vy, changes during the motion due to the acceleration of gravity. To find the initial vertical velocity, we can use the equation:
vy = v * sin(θ)
vy = 47 * sin(35°)
vy ≈ 26.87 m/s
The time taken to reach the maximum height, t, can be found using the equation:
t = vy / g
t = 26.87 / 9.8
t ≈ 2.744 s
Next, we can find the maximum height, h, using the equation:
h = vy * t - (1/2) * g * t^2
h = 26.87 * 2.744 - (1/2) * 9.8 * (2.744)^2
h ≈ 36.92 m
Therefore, the arrow will reach a maximum height of approximately 36.92 meters.
To find the maximum height reached by the arrow, we can use the equations of motion for projectile motion. The motion of the arrow can be divided into its horizontal and vertical components.
First, let's analyze the vertical component of the projectile motion. The arrow is shot at an angle of 35 degrees with the horizontal, so we need to find the vertical velocity component (Vy) of the arrow.
Vy = V * sin(θ)
Vy = 47 m/s * sin(35 degrees)
Vy ≈ 26.6 m/s
Once we have the vertical velocity component, we can find the maximum height reached by the arrow using the equation:
h = (Vy^2) / (2 * g)
where g is the acceleration due to gravity (9.8 m/s^2).
h = (26.6 m/s)^2 / (2 * 9.8 m/s^2)
h ≈ 36.25 m
Therefore, the arrow will go approximately 36.25 meters high.