A projectile is fired from the surface of the Earth with a speed of 200 meters per second at an angle of 30°
above the horizontal. If the ground is level, what is the maximum height reached by the projectile?
(A) 5 m (B) 10 m (C) 500 m (D) 1,000 m (E) 2,000 m
Vo = 200m/s[30o]
Yo = 200*sin30 = 100 m/s.
Y^2 = Yo^2 + 2g*h
h = (Y^2-Yo^2)/2g
h=(0-100^2)/-20 = 500 m. g = -10 m/s^2
To find the maximum height reached by the projectile, we can use the formula for the vertical component of the projectile's motion. The formula is given by:
𝑣² = 𝑢² − 2𝑔ℎ
Where:
𝑣 = final velocity (which is 0 at the maximum height)
𝑢 = initial velocity in the vertical direction
𝑔 = acceleration due to gravity
ℎ = maximum height reached
We first need to find the initial velocity in the vertical direction (𝑢𝑣). Given that the projectile is fired at an angle of 30° above the horizontal, we can use trigonometry to find 𝑢𝑣.
𝑢𝑣 = 𝑢 × sin(30°)
𝑢 = 200 m/s (given)
𝑢𝑣 = 200 m/s × sin(30°)
≈ 100 m/s
Using this value of 𝑢𝑣, we can find the maximum height (ℎ):
0 = (100 m/s)² − 2 × 9.8 m/s² × ℎ
0 = 10000 m²/s² − 19.6 m/s² × ℎ
19.6 m/s² × ℎ = 10000 m²/s²
ℎ = (10000 m²/s²) / 19.6 m/s²
≈ 510.2 m
Therefore, the maximum height reached by the projectile is approximately 510.2 meters. Rounding this to the nearest option, the answer is (C) 500 m.
To find the maximum height reached by the projectile, we need to analyze its motion along the vertical axis.
We know that the projectile was launched at an angle of 30° above the horizontal. Let's break down the initial velocity into its horizontal and vertical components.
The horizontal component of the velocity (Vx) can be found using trigonometry:
Vx = V * cos(30°)
= 200 m/s * cos(30°)
≈ 173.2 m/s
The vertical component of the velocity (Vy) can also be found using trigonometry:
Vy = V * sin(30°)
= 200 m/s * sin(30°)
≈ 100 m/s
Now, we can analyze the motion of the projectile along the vertical axis. The initial vertical velocity (Vy) is 100 m/s, and the only force acting on the projectile in the vertical direction is gravity.
Using the kinematic equation for vertical motion:
Vy^2 = Vo^2 + 2 * a * d
where:
Vy = final vertical velocity (0 m/s at the maximum height, as the projectile momentarily comes to a stop)
Vo = initial vertical velocity (100 m/s)
a = acceleration (due to gravity, which is approximately -9.8 m/s^2 on Earth)
d = vertical displacement (maximum height reached, which we need to find)
Rearranging the equation, we get:
d = (Vy^2 - Vo^2) / (2 * a)
Plugging in the values, we have:
d = (0^2 - 100^2) / (2 * -9.8)
= (0 - 10000) / (-19.6)
≈ 510.2 m
Therefore, the maximum height reached by the projectile is approximately 510.2 meters.
Looking at the answer choices, the closest value is 500 meters, so the correct answer is (C) 500 m.