A toy cannon fires a .11 kg shell with initial velocity vi=10m/s in the direction 52 degrees above the horizontal. The shell's trajectory curves downward because of gravity, so that time t=.61 s the shell is below the straight line by some vertical distance deltaH. find the distance "deltaH" in the absence of air resistance. the acceleration of gravity is 9.8 m/s^2. answers in units of m
What straight line are you measuring from?
y = y0 + 10*sin52*t -(g/2)*t^2
"DeltaH" = y0 - y
(assuming it is measured positive downwards)
Plug in for t and g and compute deltaH
To find the vertical distance deltaH, we can use the kinematic equation for vertical motion. The equation is given by:
Δy = viy * t + (1/2) * a * t^2
Where:
Δy is the vertical displacement (deltaH in this case)
viy is the initial vertical velocity (vi * sin θ)
t is the time of flight
a is the acceleration due to gravity (9.8 m/s^2)
Since we have the values for viy, t, and a, we can substitute them into the equation to calculate deltaH.
viy = vi * sin θ = 10 m/s * sin 52° ≈ 7.61 m/s
Δy = 7.61 m/s * 0.61 s + (1/2) * 9.8 m/s^2 * (0.61 s)^2
Δy = 4.648 m + 1.807 m
Δy ≈ 6.455 m
Therefore, the vertical distance deltaH is approximately 6.455 meters in the absence of air resistance.