What is the maximum height above ground a projectile of mass 0.67 kg, launched from ground level, can achieve if you are able to give it an initial speed of 78.5 m/s?
V^2 = Vo^2 + 2g*h
V = 0
Vo = 78.5 m/s.
g = -9.8 m/s^2.
Solve for h.
To determine the maximum height a projectile can reach, we can use the concept of projectile motion and the conservation of energy.
The first step is to find the vertical component of the initial velocity. The initial velocity can be split into horizontal and vertical components. Since the projectile is launched from ground level, the initial vertical velocity component is equal to the initial speed.
Vertical initial velocity (Vy) = 78.5 m/s
Next, we need to find the time it takes for the projectile to reach its maximum height. At the maximum height, the vertical component of the projectile's velocity (Vy) becomes zero. We can use the equation for vertical velocity in projectile motion to find the time at which this occurs:
Vy = Vy0 + gt
Where Vy0 is the initial vertical velocity, g is the acceleration due to gravity, and t is time. Rearranging the equation, we have:
g * t = -Vy0
Solving for t:
t = -Vy0 / g
Given that the acceleration due to gravity is approximately 9.8 m/s², we can substitute the values:
t = -78.5 m/s / (-9.8 m/s²)
t ≈ 8.02 seconds
Now that we have the time it takes for the projectile to reach its maximum height, we can determine the maximum height using the equation for vertical displacement in projectile motion:
Δy = Vy0 * t + (1/2) * g * t²
Substituting the values we have:
Δy = 78.5 m/s * 8.02 s + (1/2) * 9.8 m/s² * (8.02 s)²
Δy ≈ 314.91 meters
Therefore, the maximum height above the ground that the projectile can achieve is approximately 314.91 meters.