A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5 degree above the horizontal on a long flat firing range. Determine

(a) The maximum height reached by the projectile,
(b) The total time in the air,
(c) The total horizontal distance covered (that is, the range), and
(d) The velocity of the projectile 1.50 s after firing.

(a) The maximum height reached by the projectile is the vertical component of its motion. We can use the equation:

Vf^2 = Vi^2 + 2ad
where Vf is the final velocity (0 m/s at maximum height), Vi is the initial velocity (65.2 m/s), a is the acceleration (-9.8 m/s^2 due to gravity), and d is the distance traveled vertically (unknown).

Rearranging the equation to solve for d, we have:
d = (Vf^2 - Vi^2) / (2a)
d = (0^2 - 65.2^2) / (2 * -9.8)
d = -4252.8 / -19.6
d ≈ 217 meters

So, the maximum height reached by the projectile is approximately 217 meters.

(b) The total time in the air can be found by calculating the time it takes for the projectile to reach its maximum height, and then doubling that time.
The time it takes for the projectile to reach its maximum height can be found using the equation:
Vf = Vi + at
where Vf is the final velocity (0 m/s at maximum height), Vi is the initial velocity (65.2 m/s), a is the acceleration (-9.8 m/s^2 due to gravity), and t is the time taken.

Rearranging the equation to solve for t, we have:
t = (Vf - Vi) / a
t = (0 - 65.2) / -9.8
t ≈ 6.65 seconds

Since the time it takes for the projectile to reach its maximum height is approximately 6.65 seconds, the total time in the air is approximately 2 * 6.65 = 13.3 seconds.

(c) The total horizontal distance covered can be found using the equation:
d = Vi * t * cosθ
where d is the horizontal distance traveled, Vi is the initial velocity (65.2 m/s), t is the total time in the air (13.3 seconds), and θ is the launch angle (34.5 degrees).

Plugging in the values, we have:
d = 65.2 * 13.3 * cos(34.5)
d ≈ 911 meters

So, the total horizontal distance covered by the projectile is approximately 911 meters.

(d) The velocity of the projectile 1.50 seconds after firing can be found by breaking down the initial velocity into its horizontal and vertical components.
The horizontal velocity remains constant throughout the motion and can be found using the equation:
Vx = Vi * cosθ
where Vx is the horizontal component of the velocity, Vi is the initial velocity (65.2 m/s), and θ is the launch angle (34.5 degrees).

Plugging in the values, we have:
Vx = 65.2 * cos(34.5)
Vx ≈ 54.8 m/s

To find the vertical velocity at 1.50 seconds, we can use the equation:
Vy = Vi * sinθ + a * t
where Vy is the vertical component of the velocity, Vi is the initial velocity (65.2 m/s), θ is the launch angle (34.5 degrees), a is the acceleration (-9.8 m/s^2 due to gravity), and t is the time elapsed (1.50 seconds).

Plugging in the values, we have:
Vy = 65.2 * sin(34.5) + (-9.8) * 1.50
Vy ≈ 23.2 m/s

Therefore, the velocity of the projectile 1.50 seconds after firing is approximately 54.8 m/s horizontally and 23.2 m/s vertically.

To solve this problem, we can use the equations of motion for projectile motion.

(a) The maximum height reached by the projectile:

We can use the equation for vertical displacement to find the maximum height:
𝑦 = 𝑣₀𝑦𝑡 − (1/2)𝑔𝑡²

where
𝑦 = vertical displacement (maximum height)
𝑣₀𝑦 = initial vertical velocity = 𝑣₀𝑠𝑖𝑛𝑒(𝜃)
𝑔 = acceleration due to gravity
𝑡 = time

We can find the initial vertical velocity using trigonometry:
𝑣₀𝑠𝑖𝑛𝑒(𝜃) = 𝑣₀ × sin(𝜃)

Given:
𝑣₀ = 65.2 m/s
𝜃 = 34.5°
𝑔 = 9.8 m/s²

Substituting the values into the equation:
𝑦 = (65.2 m/s × sin(34.5°))𝑡 − (1/2)(9.8 m/s²)(𝑡²)

To find the maximum height, we need to find the time when the projectile reaches its maximum height. At the maximum height, the vertical velocity becomes zero (𝑣𝑦 = 0).

Using the equation 𝑣𝑦 = 𝑣₀𝑦 − 𝑔𝑡, we can find the time:
0 = (65.2 m/s × sin(34.5°)) − (9.8 m/s²)(𝑡)
𝑡 = (65.2 m/s × sin(34.5°)) / 9.8 m/s²

Now, substitute the value of 𝑡 back into the initial equation to find the maximum height.

(b) The total time in the air:

The total time in the air is twice the time it takes for the projectile to reach its maximum height.
𝑡_total = 2(𝑡)

(c) The total horizontal distance covered (range):

The horizontal distance covered (range) can be found using the equation:
𝑥 = 𝑣₀𝑥 𝑡

To calculate 𝑣₀𝑥, we can use the equation 𝑣₀𝑥 = 𝑣₀ × cos(𝜃).

(d) The velocity of the projectile 1.50 s after firing:

We can use the equations of motion to find the horizontal and vertical components of the velocity at a given time 𝑡.

Horizontal velocity:
𝑣𝑥 = 𝑣₀𝑥

Vertical velocity:
𝑣𝑦 = 𝑣₀𝑦 − 𝑔𝑡

The overall velocity 𝑣 can be calculated using the Pythagorean theorem:
𝑣 = √(𝑣𝑥² + 𝑣𝑦²)

Substituting the given time into these equations will give us the answer.

Please provide any value of 𝑡 or specific part (a, b, c, or d) you would like me to calculate.

To solve these problems, we can use the equations of projectile motion.

(a) To find the maximum height reached by the projectile, we need to find the vertical component of the initial velocity. We can use the equation:
Vy = V * sin(θ),
where Vy is the vertical component of the initial velocity, V is the initial speed, and θ is the angle above the horizontal.
Substituting the given values, we have:
Vy = 65.2 m/s * sin(34.5°) ≈ 35.18 m/s.

The maximum height can be determined using the equation for vertical displacement:
Δy = (Vy^2) / (2 * g),
where Δy is the vertical displacement and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values, we get:
Δy = (35.18 m/s)^2 / (2 * 9.8 m/s^2) ≈ 63.09 m.

Therefore, the maximum height reached by the projectile is approximately 63.09 m.

(b) To find the total time in the air, we can use the equation:
t = 2 * (Vy / g),
where t is the total time of flight.
Substituting the values, we get:
t = 2 * (35.18 m/s) / (9.8 m/s^2) ≈ 7.20 s.

Therefore, the total time in the air is approximately 7.20 seconds.

(c) To find the total horizontal distance covered (range), we can use the equation:
R = Vx * t,
where R is the range, Vx is the horizontal component of the initial velocity, and t is the total time of flight.
The horizontal component of the initial velocity can be found using the equation:
Vx = V * cos(θ).
Substituting the values, we get:
Vx = 65.2 m/s * cos(34.5°) ≈ 54.13 m/s.

Now we can calculate the range:
R = (54.13 m/s) * (7.20 s) ≈ 389.95 m.

Therefore, the total horizontal distance covered (range) is approximately 389.95 m.

(d) To find the velocity of the projectile 1.50 s after firing, we can break it down into its vertical and horizontal components.
The vertical component of the velocity remains constant, and the horizontal component remains unchanged.

The vertical component of the velocity can be found using the equation:
Vy = V * sin(θ).
Substituting the values, we get:
Vy = 65.2 m/s * sin(34.5°) ≈ 35.18 m/s.

For the horizontal component of the velocity, Vx, it remains constant throughout the motion and is given by:
Vx = V * cos(θ).
Substituting the values, we get:
Vx = 65.2 m/s * cos(34.5°) ≈ 54.13 m/s.

Therefore, the velocity of the projectile 1.50 s after firing is approximately 54.13 m/s horizontally and 35.18 m/s vertically.

do vertical problem

find Vi, initial speed up
Vi = 65.2 sin 34.5

find t where v = 0, top of curve
v = Vi - 9.81 t
0 = Vi - 9.81 t
solve for t, time at top, half the time in the air

max height
h = 0 + Vi t - 4.9 t^2

now horizontal problem
u = 65.2 * cos 34.5 the whole time
range = u*time in air = u (2 t)

(d)
horizontal component = u the whole time
vertical component = v = Vi - 9.81 (1.5)

speed = sqrt (u^2 + v^2)
tan angle to horizontal = v/u