You have an spring gun. The end of the gun is at 1 meter above the ground. This gun is shoot horizontally with no air resistance. The cannon ball hits the ground 1.32 meters ahead from it was shoot. What is the cannon ball's initial velocity?
h = 0.5g*t^2 = 1 m.
4.9t^2 = 1
t^2 = 0.204
t = o.452 s. = Fall time.
d = Vo*t = 1.32 m.
Vo*0.452 = 1.32
Vo = 2.92 m/s.
To find the initial velocity of the cannon ball, we can use the equation of motion for horizontal projectile motion:
x = vt
Where:
x is the horizontal distance traveled by the cannonball (1.32 meters)
v is the initial velocity of the cannonball (what we're trying to find)
t is the time of flight (which we'll determine later)
Since the cannonball is shot horizontally, its initial vertical velocity is zero. We can use the equation of motion for vertical projectile motion to find the time of flight:
y = V₀t + 0.5gt²
Where:
y is the vertical distance traveled by the cannonball (1 meter)
V₀ is the initial vertical velocity (zero in this case)
g is the acceleration due to gravity (approximately 9.8 m/s²)
t is the time of flight (what we're trying to find)
Plugging in the values we know:
1 = 0.5 * 9.8 * t²
Simplifying the equation:
1 = 4.9t²
Now we can solve for t by taking the square root of both sides:
t = √(1/4.9) ≈ 0.451 seconds
Now that we have the time of flight, we can substitute it back into the equation for horizontal motion:
1.32 = v * 0.451
Solving for v:
v = 1.32 / 0.451 ≈ 2.92 m/s
Therefore, the cannonball's initial velocity is approximately 2.92 m/s.