A cannon fires a 0.598 kg shell with initial velocity 10 m/s in the direction 46◦ above the horizontal.
The shell’s trajectory curves downward be- cause of gravity, so at the time t = 0.53 s the shell is below the straight line by some verti- cal distance ∆h. Your task is to calculate the distance ∆h in the absence of air resistance.
On what does ∆h depend (besides g)?
That didn't answer the question
To calculate the vertical distance ∆h in the absence of air resistance, we need to consider the projectile motion of the shell. This involves using the equations of motion and taking into account the initial velocity, launch angle, and time.
The equation for the vertical position of a projectile as a function of time is given by:
y(t) = y₀ + v₀y * t - (1/2) * g * t²,
where:
- y(t) is the vertical position of the projectile as a function of time,
- y₀ is the initial vertical position (in this case, ∆h),
- v₀y is the initial vertical component of velocity (v₀y = v₀ * sin(θ)),
- t is the time elapsed, and
- g is the acceleration due to gravity (approximately 9.8 m/s²).
From the problem statement, we are given:
- mass of the shell (m) = 0.598 kg,
- initial velocity of the shell (v₀) = 10 m/s, and
- launch angle (θ) = 46° above the horizontal.
To find ∆h, we need to calculate the vertical position of the shell at t = 0.53 s.
Let's calculate it step by step:
1. Calculate the initial vertical component of velocity:
v₀y = v₀ * sin(θ)
= 10 m/s * sin(46°)
2. Substitute the given values into the equation for y(t):
y(t) = ∆h + (v₀ * sin(θ)) * t - (1/2) * g * t²
3. Substitute t = 0.53 s into the equation:
y(0.53 s) = ∆h + (v₀ * sin(θ)) * 0.53 s - (1/2) * g * (0.53 s)²
4. Simplify and calculate the result to find ∆h.
Therefore, besides gravity (g), the vertical distance ∆h depends on the initial velocity (v₀), the launch angle (θ), and the time (t).
Problem entirely vertical assuming your straight line is horizontal
Vi = 10 sin 46
v = Vi - g t
delta h = Vi t - (1/2) g t^2