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, we can use the equations of motion for projectile motion. The vertical motion and horizontal motion are independent of each other.
First, let's find the time taken by the projectile to reach its maximum height.
The initial vertical velocity (Vy) can be calculated using the initial speed (v) and the launch angle (θ):
Vy = v * sin(θ)
Vy = 200 * sin(30°)
Vy = 100 m/s
Next, we can use the equation for vertical motion to find the time taken to reach the maximum height (t):
Vy = u + at
Where:
Vy = final vertical velocity (0 m/s at the maximum height)
u = initial vertical velocity
a = acceleration due to gravity (-9.8 m/s^2)
0 = 100 - 9.8t
9.8t = 100
t = 100 / 9.8
t ≈ 10.2 seconds
Now, we can use the equation for vertical displacement to find the maximum height (h):
h = u * t + (1/2) * a * t^2
Where:
u = initial vertical velocity
t = time taken to reach the maximum height
a = acceleration due to gravity
h = 100 * 10.2 + (1/2) * (-9.8) * (10.2)^2
h = 1020 - 504.36
h ≈ 515 m
Therefore, the maximum height reached by the projectile is approximately 515 meters.
Since none of the available answer choices exactly match 515 meters, you can round it to the nearest option. In this case, the answer would be (C) 500 m.