An arrow is shot at an angle 39◦ with the

horizontal. It has a velocity of 54 m/s.
How high will the arrow go? The acceleration of gravity is 9.8 m/s^2. Answer in units of m

To find the maximum height reached by the arrow, we can use the kinematic equations of motion.

1. First, we need to break down the initial velocity into horizontal and vertical components. The vertical component can be determined using the given angle and velocity.

Vertical Component (V_y) = Velocity * sin(angle)
V_y = 54 m/s * sin(39°)
V_y = 54 m/s * 0.6293
V_y ≈ 33.9822 m/s

2. Next, we can calculate the time (t) it takes for the arrow to reach its highest point. The time to reach the maximum height is equal to the time taken for the vertical component of the velocity to become zero.

Using the vertical motion equation:
V_fy = V_iy + gt

Since the final vertical velocity at the highest point is 0 (since the arrow momentarily stops moving upward), we can rearrange the equation to solve for t:

0 = 33.9822 m/s - (9.8 m/s^2) * t

Solving for t:
9.8t = 33.9822
t = 33.9822 / 9.8
t ≈ 3.4684 s

3. Now that we have found the time it takes for the arrow to reach its highest point, we can calculate the maximum height (H) using the vertical motion equation:

H = V_iy * t - (1/2) * g * t^2

H = 33.9822 m/s * 3.4684 s - (1/2) * (9.8 m/s^2) * (3.4684 s)^2

H ≈ 117.536 m

Therefore, the arrow will reach a maximum height of approximately 117.536 meters.

Vi = 54 sin 39

v = Vi - 9.8 t
v = 0 at top
so
t = Vi/9.8 at top

use that t
h = Vi t - 4.9 t^2

it will turn out to be 1/2 Vi t