A military helicopter on a training mission is flying horizontally at a speed of 60.0 m/s and accidentally drops a bomb (fortunately not armed) at an elevation of 350 m. You can ignore air resistance.
Find the horizontal and vertical components of its velocity just before it strikes the earth.
a) d = 1/2(a x t^2); rewrite to solve for t:
t = sqrt(2 x d / a) = sqrt( 2 x 350 / 9.8) = 8.45 seconds
b) horiz distance = v x t = 60 x 8.45 = 507 meters
c) bombs horizontal component of velocity doesn't change, neglecting air resistance = 60.0 m/s.
d) the helicopter is directly over the bomb if no air resistance
To find the horizontal and vertical components of the helicopter's velocity just before the bomb strikes the earth, we can break down the problem into two parts: the horizontal motion and the vertical motion.
Let's start with the horizontal motion. Since the helicopter is flying horizontally, its horizontal velocity remains constant throughout. So, the horizontal component of the velocity just before the bomb strikes the earth is also 60.0 m/s.
Now, let's look at the vertical motion. We know that the initial vertical velocity of the bomb is 0 m/s since it was dropped. The only force acting on the bomb in the vertical direction is gravity, causing it to accelerate downward.
We can use the kinematic equation to find the final vertical velocity of the bomb just before it strikes the earth. The equation we'll use is:
vf^2 = vi^2 + 2ad
Where:
- vf is the final vertical velocity
- vi is the initial vertical velocity (0 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2, assuming positive direction is upward)
- d is the displacement in the vertical direction (350 m)
Plugging in the values, we have:
vf^2 = 0^2 + 2(-9.8)(350)
vf^2 = -6860
vf = sqrt(-6860)
vf = 82.8 m/s (approximately)
Therefore, the vertical component of the velocity just before the bomb strikes the earth is approximately 82.8 m/s.
To summarize:
- Horizontal component of velocity: 60.0 m/s
- Vertical component of velocity: 82.8 m/s