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
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
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.