A bullet is fired horizontally from the roof of a building 100 meters tall with a speed of 850 m/s. Neglecting air resistance, how far will the bullet drop in 3 seconds?
1 29.4 m
2 44.1 m
3 100 m
4 2550 m
Answer should be choice 2) 44.1m
BUT I DON'T GET IT!
Please help me;(
if i remember the formula correctly as:
x = (-9.8/2)t^2 + vt + x0
x = (-9.8/2)(3^2)
x = 44.1 m
You split up this problem into horizontal and vertical components like below:
a = acceleration
v = velocity
x = distance
x0 = starting distance (usually 0)
t = time
Vertical:
a = -9.8m/s^2
v = none (they said that the bullet was shot horizontally so vertically, it would be like the velocity at the peak of a free-fall.
x = 100 m
x0 = 0 m (this is like the starting point of the fall so technbically 0)
Horizontal:
a = none (gravity doesn't move sideways and there's no given acceleration)
v = 850m/s
x = (not given)
x0 = (not given)
:)
So, the equation for the vertical component is:
x = (-9.8/2)t^2 + vt + x0
x = (-9.8/2)(3^2) + 0 + 100
x = 44.1 m
Therefore, the answer is choice 2) 44.1m
Ah, I see you're having trouble understanding the problem. Let me break it down for you in a way that will hopefully make you smile!
So, we have a bullet being fired horizontally from a 100-meter tall building. It's like the bullet is saying, "I'm on top of the world!" But gravity has other plans, and it pulls the bullet downward as it travels.
We want to find out how far the bullet will drop in 3 seconds. Now, to calculate this, we need to consider the vertical motion of the bullet. The horizontal motion doesn't come into play here because the bullet is fired horizontally. It's like that bullet just wants to take a stroll in the park!
We can use the equation x = (-9.8/2)t^2 + vt + x0, where x represents the distance, t is the time, v is the initial velocity, and x0 is the starting distance.
Since the bullet is moving horizontally, its initial velocity in the vertical direction is zero. It's like saying, "No jumping for me, I'm strictly going sideways!"
Plugging in the values, we have x = (-9.8/2)(3^2) = 44.1 meters. So, after 3 seconds, the bullet will drop 44.1 meters. It's like it's doing a skydiving stunt without a parachute!
And there you have it, my friend! The answer is choice 2) 44.1 meters. I hope this explanation brought a smile to your face and cleared up any confusion. Keep up the great work!
You have correctly applied the vertical kinematic equation to solve for the distance the bullet drops in 3 seconds. The equation you used, x = (-9.8/2)t^2 + vt + x0, is the correct formula to calculate the vertical distance traveled under constant acceleration due to gravity, neglecting air resistance.
In this case, the initial vertical position (x0) is 0 since the bullet starts from the roof of the building, the initial vertical velocity (v) is also 0 because the bullet is fired horizontally, and the acceleration (a) is -9.8 m/s^2 due to gravity pulling the bullet downward.
Substituting these values into the equation:
x = (-9.8/2)(3^2) = -44.1 m
The negative sign indicates that the bullet drops downwards. However, since the question asks for the absolute value of the distance, the answer is 44.1 meters.
So, the correct answer is choice 2) 44.1 m.
To find the horizontal distance the bullet will drop, we can use the horizontal components of the bullet's motion. Since the bullet is fired horizontally, there is no horizontal acceleration or initial horizontal position.
We are given the initial horizontal velocity of the bullet (850 m/s), and we need to find the horizontal distance traveled in 3 seconds.
The formula for horizontal distance is:
x = v * t
where:
x = horizontal distance
v = initial horizontal velocity
t = time
Plugging in the given values:
x = 850 m/s * 3 s
x = 2550 m
Therefore, the horizontal distance the bullet will drop in 3 seconds is 2550 meters.