If a shot is fired directly upward at 1200 feet per second from a ground elevation of 8oo feet above sea level, how high would it go above the ground level?
Thanks,
Gerry
To determine the height the shot would reach above the ground level, we can use kinematic equations of motion. In particular, we can use the equation for vertical displacement:
Δy = v₀t + (1/2)gt²
Where:
Δy is the vertical displacement or height,
v₀ is the initial velocity,
t is the time, and
g is the acceleration due to gravity.
In this case, the initial velocity is given as 1200 feet per second. However, we need to take into account that the shot is fired from a ground elevation of 800 feet above sea level. So, effectively, the initial height would be 800 feet.
We know that the shot will travel upwards until it reaches the peak of its trajectory, where the vertical displacement will be the highest. At this point, the final velocity will be 0, and the acceleration due to gravity will be acting in the opposite direction. We can find the time it takes to reach the peak using the following equation:
0 = v₀ - gt
Rearranging the equation, we have:
t = v₀ / g
Substituting the values, we get:
t = 1200 ft/s / 32.2 ft/s²
Simplifying, we find:
t ≈ 37.27 s
Now, we can determine the vertical displacement or height using the equation Δy = v₀t + (1/2)gt².
Δy = (1200 ft/s)(37.27 s) + (1/2)(32.2 ft/s²)(37.27 s)²
Calculating, we find:
Δy ≈ 446,695.16 ft
Therefore, the shot would reach approximately 446,695.16 feet above the ground level.