A bullet is fired horizontally from the top of a cliff which is 750 meters high and hits the ground at a horizontal distance of 1660 meters. (Consider air resistance to be negligible
To find the initial velocity of the bullet, we can use the fact that the vertical displacement is equal to the initial vertical velocity multiplied by the time taken to reach the ground.
Given:
- Height of the cliff (h) = 750 meters
- Horizontal distance traveled (d) = 1660 meters
- Vertical displacement (Δy) = -750 meters (negative since the displacement is downward)
First, we need to find the time taken to reach the ground. We can use the equation of motion for vertical motion:
Δy = (1/2) * g * t^2
Where:
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken to reach the ground
Substituting the known values:
-750 = (1/2) * 9.8 * t^2
Dividing both sides by (1/2) * 9.8:
t^2 = -750 / (1/2 * 9.8) = -750 / 4.9
Taking the square root of both sides (we ignore the negative root since time cannot be negative):
t ≈ sqrt(-750 / 4.9)
Next, we can use the equation of motion for horizontal motion to find the initial velocity of the bullet:
d = v_initial * t
Rearranging the equation:
v_initial = d / t
Substituting the known values:
v_initial = 1660 / t
Calculating the value of t using the previous result:
t ≈ sqrt(-750 / 4.9) ≈ 12.198 seconds
Substituting back into the equation:
v_initial = 1660 / 12.198
Calculating:
v_initial ≈ 136 meters per second
Therefore, the initial velocity of the bullet is approximately 136 meters per second.