A projectile is launched at an angle of 20.0 degrees above the horizontal with an initial speed of 88.1 m/s. The maximum height reached by the projectile will be ____ m.
Vi = 55.1 sin 20
v = Vi - 9.81 t
v = 0 at top
so
t = Vi/9.81 at top
use that in
h = Vi t - 4.9 t^2
To determine the maximum height reached by the projectile, we need to use kinematic equations of motion.
First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component does not change throughout the trajectory, so we can define it as:
Vx = initial velocity * cos(angle)
where:
Vx = horizontal component of velocity
angle = launch angle in degrees
Given that the initial velocity is 88.1 m/s and the launch angle is 20.0 degrees, we can calculate the horizontal component of velocity:
Vx = 88.1 m/s * cos(20.0 degrees)
Vx ≈ 83.93 m/s
The vertical component of the initial velocity can be calculated as:
Vy = initial velocity * sin(angle)
Vy = 88.1 m/s * sin(20.0 degrees)
Vy ≈ 30.11 m/s
Now, let's determine the time it takes for the projectile to reach its highest point. At the maximum height, Vy is equal to 0 because the projectile momentarily stops moving upwards. We can use the following equation to find the time of flight:
Vy = initial velocity * sin(angle) - g * t
Since Vy becomes 0, we can rearrange the equation to solve for time (t):
0 = 88.1 m/s * sin(20.0 degrees) - 9.8 m/s^2 * t
Solving for t, we get:
t = 30.11 m/s / 9.8 m/s^2
t ≈ 3.07 seconds
Now that we know the time of flight, we can find the maximum height reached by the projectile using the vertical component of the displacement formula:
Δy = Vy * t - 0.5 * g * t^2
where:
Δy = vertical displacement (maximum height)
Vy = vertical component of initial velocity
t = time of flight
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values:
Δy = 30.11 m/s * 3.07 s - 0.5 * 9.8 m/s^2 * (3.07 s)^2
Δy ≈ 45.64 m
Therefore, the maximum height reached by the projectile will be approximately 45.64 meters.