A model rocket is launched at an angle of 80 degrees aboce the x axis, with a velocity of 30m/s. How high will the rocket be 5 seconds after launch?

ROund your answer to the nearest tenth

Vo = 30m/s[80o]

Yo = 30*sin80 = 29.54 m/s.

hmax = (V^2-Vo^2)/2g
hmax = (0-29.54^2)/-19.6 = 44.52 m
Tr = -Vo/g = -29.54/-9.8 = 3.0 s = Rise time,

Tf = 5-3 = 2 s. = Fall time.

h = ho - 0.5g*t^2
h=44.52 - 4.9*2^2 = 24.92 m. after 5 s.

4.9*2^2 = 19.6

To find the height of the rocket 5 seconds after launch, we can use the following steps:

Step 1: Split the initial velocity into its horizontal and vertical components:
The initial velocity (v₀) can be split into its horizontal (v₀x) and vertical (v₀y) components using trigonometry.
v₀x = v₀ * cosθ
v₀y = v₀ * sinθ

Given:
v₀ = 30 m/s (initial velocity)
θ = 80 degrees (angle above the x-axis)

Step 2: Calculate the vertical displacement (change in height):
The vertical displacement can be calculated using the following equation:
y = v₀y * t + (1/2) * a * t²
where:
y = vertical displacement (height)
v₀y = vertical component of initial velocity
t = time (5 seconds)
a = acceleration due to gravity (approximately -9.8 m/s²)

Step 3: Substitute the known values into the equation and calculate the height:
y = (v₀ * sinθ) * t + (1/2) * (-9.8) * t²
y = 30 * sin(80°) * 5 + (1/2) * (-9.8) * (5)²
y ≈ 10.074 + (-122.5)
y ≈ -112.426

Since height cannot be negative, we round the answer to the nearest tenth:
The rocket will be approximately 112.4 meters high 5 seconds after launch.

To find the height of the rocket 5 seconds after launch, we can break down the motion into horizontal and vertical components.

First, let's analyze the vertical motion of the rocket. We know the initial vertical velocity (Viy) is given by Viy = V * sin(θ), where V is the launch velocity (30 m/s) and θ is the launch angle (80 degrees).

Plugging in the values, we have Viy = 30 m/s * sin(80°) = 29.9432 m/s (rounded to four decimal places).

Using the formula for vertical displacement (y), we can calculate the height (h) of the rocket after 5 seconds. The formula is:

y = Viy * t + (1/2) * g * t^2,

where t is the time (5 seconds) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the given values, we have:

y = (29.9432 m/s) * (5 s) + (1/2) * (9.8 m/s^2) * (5 s)^2
= 149.716 m + 122.5 m
= 272.216 m (rounded to three decimal places).

Therefore, the height of the rocket 5 seconds after launch is approximately 272.2 meters.