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
See previous post.
To find the maximum height reached by the projectile, you can use the following steps:
Step 1: Split the initial velocity into its vertical and horizontal components:
The vertical component of the initial velocity (v0) can be found by multiplying the magnitude of the velocity (200 meters per second) by the sine of the launch angle (30 degrees):
Vertical component (v0y) = v0 * sin(angle)
Step 2: Calculate the time it takes for the projectile to reach its maximum height:
The time taken to reach maximum height (t_max) can be found by dividing the vertical component of the initial velocity (v0y) by the acceleration due to gravity (g) which is approximately 9.8 meters per second squared.
t_max = v0y / g
Step 3: Calculate the maximum height (h_max) reached by the projectile:
The maximum height can be determined by using the formula for the height of an object in projectile motion:
h_max = v0y^2 / (2 * g)
Plugging in the values:
v0 = 200 meters per second
angle = 30 degrees
g = 9.8 meters per second squared
Step 1:
v0y = 200 * sin(30) ≈ 100 meters per second
Step 2:
t_max = 100 / 9.8 ≈ 10.2 seconds (rounded to one decimal place)
Step 3:
h_max = (100^2) / (2 * 9.8) ≈ 510 meters (rounded to the nearest meter)
Therefore, the maximum height reached by the projectile is approximately 510 meters.
The closest option to this value is (C) 500 m.