a rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 75m/s. It moves for 25 seconds along its line of motion with an overall acceleration of 25 m/s^2. What is its maximum height?

Answer 1

First, let us find the vertical and horizontal components of the initial velocity.

The vertical component of the initial velocity (Vv0) can be calculated as follows:
Vv0 = Vi * sin(θ)
where:
Vi = initial speed = 75 m/s
θ = angle of projection = 53°

Vv0 = 75 * sin(53°)
Vv0 ≈ 60 m/s

Next, let's find the acceleration. The given acceleration is along the line of motion at the 53-degree angle, so we have to find the vertical component of the acceleration.

The vertical component of the acceleration (Av) can be calculated as follows:
Av = Atotal * sin(θ)
where:
Atotal = overall acceleration = 25 m/s²
θ = angle of projection = 53°

Av = 25 * sin(53°)
Av ≈ 20 m/s²

Now we have the initial vertical velocity Vv0 and vertical acceleration Av. We can find the time taken to reach the maximum height.
At the maximum height, the vertical velocity will be 0. We can use the following formula to find the time taken:

Vvf = Vv0 + a * t
where:
Vvf = final vertical velocity = 0
Vv0 = initial vertical velocity = 60 m/s
a = acceleration = -20 m/s² (negative value because it opposes the motion)
t = time taken to reach maximum height

Solving for t:
0 = 60 - 20 * t
t = 60 / 20
t = 3 seconds

Now that we have the time taken to reach the maximum height, we can find the maximum height using the following formula:

h = Vv0 * t + 0.5 * a * t²
where:
Vv0 = initial vertical velocity = 60 m/s
t = time taken to reach maximum height = 3 seconds
a = acceleration = -20 m/s²
h = maximum height

h = 60 * 3 + 0.5 * (-20) * 3²
h = 180 - 0.5 * 20 * 9
h = 180 - 90
h = 90 meters

The rocket's maximum height is 90 meters.