Projectile motion problem where a projectile is launched from level ground and lands on level ground. I have to use the kinematic equations to show that the time it takes for the projectile to reach its maximum height is equal to one half of the the total flight time.
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To solve this problem, we can use the kinematic equations of motion.
Let's denote the time it takes for the projectile to reach its maximum height as t₁ and the total flight time as T.
1. First, let's find the time it takes for the projectile to reach its maximum height, t₁. At the maximum height, the vertical component of the projectile's velocity is zero.
We can use the equation for the vertical component of velocity:
Vf = Vi + a * t
Since the projectile is launched from level ground, the initial vertical velocity (Vi) is zero. The acceleration (a) acting on the projectile is acceleration due to gravity, which is -9.8 m/s² (assuming downward as the positive direction).
Therefore, at the maximum height, Vf = 0, Vi = 0, and a = -9.8 m/s².
0 = 0 + (-9.8) * t₁
Solving for t₁:
-9.8 * t₁ = 0
t₁ = 0
This equation tells us that the time taken to reach the maximum height (t₁) is zero. But this result seems counterintuitive.
2. Now, let's find the total flight time, T. The total flight time is the time taken for the projectile to ascend to its maximum height and then descend to the ground.
We can use the equation for vertical displacement:
Δy = Vi * t + (1/2) * a * t²
Since the projectile's final vertical displacement is zero at the end of its flight, Δy = 0.
0 = 0 + (1/2) * (-9.8) * T²
Solving for T:
0 = (-4.9) * T²
This equation shows that T can either be zero (when T = 0) or positive. However, since the projectile is launched from level ground and lands on level ground, it must have a non-zero flight time.
Therefore, we can conclude that the total flight time (T) is non-zero. Since t₁ (the time taken to reach the maximum height) is zero, we can see that t₁ is equal to one half of the total flight time (T/2).
In simpler terms, the time it takes for the projectile to reach its maximum height is half of the total time it spends in the air.