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?

SIMPLIFY.

To solve part (a) of the problem and determine the maximum height the arrow will attain, we need to analyze the projectile motion of the arrow. We can break down the initial velocity of the arrow into its horizontal and vertical components.

Given:
Initial speed (v₀) = 54 m/s
Launch angle (θ) = 29.0°

Step 1: Resolve the initial velocity into horizontal and vertical components.
The horizontal component (v₀x) remains unchanged throughout the motion, while the vertical component (v₀y) changes due to gravity.

v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)

Step 2: Determine the time it takes for the arrow to reach its maximum height.
At the highest point of flight, the vertical velocity becomes zero. Using this information, we can calculate the time it takes to reach this point (t_max).

v_y = v₀y - g * t
0 = v₀y - g * t_max

Solve for t_max:
t_max = v₀y / g

Step 3: Calculate the maximum height (h_max).
To find the maximum height, we need to figure out the vertical displacement traveled by the arrow. We can use the equation:

h = v₀y * t - 0.5 * g * t^2

Since the arrow reaches its maximum height at t_max, we plug in the value for t_max:

h_max = v₀y * t_max - 0.5 * g * t_max^2

Now, let's calculate the values.

Substituting the known values:
v₀ = 54 m/s
θ = 29.0°
g = 9.8 m/s²

v₀x = 54 m/s * cos(29.0°)
v₀y = 54 m/s * sin(29.0°)

t_max = v₀y / g
h_max = v₀y * t_max - 0.5 * g * t_max^2

After solving these equations, we can find the maximum height attained by the arrow (h_max).

To solve part (b) of the problem and determine how far the target is, we need to analyze the horizontal motion of the arrow.

Step 4: Calculate the total time of flight (t_total).
In projectile motion, the total time of flight is when the object returns to the same vertical level from which it was launched. Since the target is at the same height from which the arrow was shot, the total time of flight is twice the time taken to reach the maximum height.

t_total = 2 * t_max

Step 5: Calculate the horizontal displacement traveled by the arrow (range).
The horizontal displacement can be calculated using the equation:

range = v₀x * t_total

Now, let's calculate the values.

Substituting the known values:
v₀x = 54 m/s * cos(29.0°)
t_total = 2 * t_max

After solving this equation, we can find the horizontal displacement traveled by the arrow, which represents the distance to the target (range).

at the max height, velocity vertical is zero.

Vf=0=54*sin29deg - g*t
solve for t. then put it in this:
hmax=54*sin29*t-1/2 g t^2

distance target:= 54cos29*(2t) where t is the time to the max height, 2t is the total time in the air.