2. A bullet is fired from the ground at an angle of 45o above the horizontal. What initial speed vo must the bullet have in order to hit a point 550 ft high on a tower located 600 ft away (ignoring air resistance)?
Y^2 = Yo^2 + 2g*d = 0.
Yo^2 - 64*550 = 0.
Yo^2 = 64*550 = 35,200.
Yo = 187.6 Ft/s = Ver. component of Vo.
Vo = Yo/sinA=187.6 / sin45=265.3 Ft/s. = Initial velocity.
To solve this problem, we can break it down into two components: horizontal and vertical motion.
First, let's analyze the vertical motion of the bullet. We know that the initial vertical velocity (vy) is determined by the initial speed (vo) and the launch angle (θ). Since the bullet is launched at an angle of 45° above the horizontal, the vertical component is vy = vo * sin(45°).
Next, we need to determine how long it takes for the bullet to reach its maximum height. At the highest point, the vertical velocity will be zero. We can use this information to find the time taken for the bullet to reach its peak.
The vertical displacement (Δy) is given as 550 ft. We can use the following kinematic equation:
Δy = vy * t + 0.5 * g * t^2
Since the bullet reaches its highest point, the final vertical displacement is zero. Hence,
0 = vy * t + 0.5 * g * t^2
Now we can substitute the value of vy we found earlier:
0 = (vo * sin(45°)) * t + 0.5 * g * t^2
Simplifying further:
t = (vo * sin(45°)) / g
Now, let's analyze the horizontal motion of the bullet. The horizontal motion is unaffected by gravity, and the horizontal displacement (Δx) is given as 600 ft. The time taken for the horizontal motion is the same time we obtained earlier: t.
Now we can determine the horizontal velocity (vx) using the formula:
Δx = vx * t
Substituting the values:
600 ft = vx * t
Now, let's express the horizontal velocity in terms of vo and θ. The horizontal component of the velocity (vx) can be given as vx = vo * cos(45°).
Substituting the values again:
600 ft = (vo * cos(45°)) * t
Now, substitute the value of t we found earlier:
600 ft = (vo * cos(45°)) * [(vo * sin(45°)) / g]
Simplifying further:
600 ft = [(vo)^2 * cos(45°) * sin(45°)] / g
Solving for vo:
vo^2 = (600 ft * g) / [cos(45°) * sin(45°)]
Finally, taking the square root of both sides:
vo = √[(600 ft * g) / [cos(45°) * sin(45°)]]
Now you can plug in the value of acceleration due to gravity (g ≈ 32.2 ft/s²) and evaluate the expression to find the initial speed (vo) in order to hit the point 550 ft high on the tower located 600 ft away.