A 1.20- block of wood sits at the edge of a table, 0.720 above the floor. A 1.20×10−2- bullet moving horizontally with a speed of 750 embeds itself within the block. What horizontal distance does the block cover before hitting the ground?
Well, let me see. If we're talking about a bullet embedding itself in a block of wood, I hope it's not because they got into an argument. That's definitely not the best way to resolve conflicts.
Now, to figure out the horizontal distance the block covers before hitting the ground, we need to consider a couple of things. First, we need to find out how much time it takes for the block to hit the ground.
Since the bullet embeds itself in the block, we can assume that the bullet's horizontal velocity is transferred to the block. That means the block will have an initial horizontal velocity of 750 m/s.
Now, let's find out how much time it takes for the block to hit the ground. We can use the equation 𝑑 = 𝑣𝑖𝑡 + 0.5𝑎𝑡^2, where 𝑑 is the vertical distance, 𝑣𝑖 is the initial vertical velocity, 𝑡 is the time, and 𝑎 is the acceleration.
Since the block is falling under gravity, the acceleration is 9.8 m/s^2. The initial vertical velocity is 0 because the block is initially at rest. The vertical distance is given as 0.720 m.
Plugging in the values, we can solve for 𝑡. Let me just do a quick calculation... *beep boop beep* Alright, it takes approximately 0.382 seconds for the block to hit the ground.
Now, we can find the horizontal distance covered by multiplying the time by the initial horizontal velocity. That will give us (0.382 s) * (750 m/s) = 286.5 meters.
Therefore, the block covers approximately 286.5 meters horizontally before hitting the ground. Just a friendly reminder though, it's always good to stay away from situations where bullets and blocks of wood are involved. Safety first!
To find the horizontal distance the block covers before hitting the ground, we can use the concept of projectile motion.
Step 1: Determine the time it takes for the block to hit the ground.
We can use the equation for vertical motion, considering the initial vertical velocity is zero, the height is 0.720 m, and the acceleration due to gravity is -9.8 m/s^2.
Using the equation:
h = v_i*t + (1/2)*a*t^2,
where h is the height, v_i is the initial vertical velocity, a is the acceleration, and t is the time, we can solve for t:
0.720 = 0*t + (1/2)*(-9.8)*t^2.
Simplifying the equation, we get:
0.720 = -4.9*t^2.
Rearranging the equation:
t^2 = 0.720 / -4.9.
Taking the square root of both sides:
t ≈ √(0.720 / -4.9).
Calculating the value of t:
t ≈ √(0.720 / -4.9) ≈ 0.383 s.
Step 2: Determine the horizontal distance traveled by the block.
Since the horizontal motion of the block is not affected by gravity, the horizontal speed remains constant at 750 m/s.
Using the formula:
d = v*t,
where d is the distance, v is the velocity, and t is the time, we can calculate the horizontal distance:
d = 750 m/s * 0.383 s.
Calculating the value of d:
d ≈ 287.25 m.
Therefore, the block covers a horizontal distance of approximately 287.25 meters before hitting the ground.
To find the horizontal distance the block covers before hitting the ground, we can use the principles of projectile motion.
First, let's determine the time it takes for the block to hit the ground. We have the height of the table above the floor, which is 0.720 m. We can use the equation for vertical motion:
h = ut + (1/2)gt^2
Where:
h = height (0.720 m)
u = initial vertical velocity (0 m/s, as the block is initially at rest)
g = acceleration due to gravity (-9.8 m/s^2, taking downward as negative)
t = time taken
Rearranging the equation, we get:
t^2 - 2(h/g) = 0
Plugging in the values, we have:
t^2 - 2(0.720 m)/(-9.8 m/s^2) = 0
Solving for t, we find:
t ≈ 0.38 s
Now, we can determine the horizontal distance covered by the block in this time. Since the horizontal velocity remains constant throughout the motion, we can use the equation:
distance = speed × time
The speed of the bullet is given as 750 m/s, and the time is 0.38 s. Plugging in the values, we have:
distance = 750 m/s × 0.38 s
Solving this, we find:
distance ≈ 285 m
Therefore, the horizontal distance the block covers before hitting the ground is approximately 285 meters.