To solve this problem, we can split the initial velocity of the shell into its horizontal and vertical components.
(a) The horizontal component of the initial velocity can be found using the formula:
v_x = v * cos(theta)
where v_x is the horizontal component, v is the muzzle speed, and theta is the elevation angle. Plugging in the given values, we have:
v_x = 75.0 m/s * cos(35°) = 61.22 m/s
Now, we can calculate the time it takes for the shell to hit the ground. The vertical component of the initial velocity can be found using the formula:
v_y = v * sin(theta)
where v_y is the vertical component. Plugging in the given values, we have:
v_y = 75.0 m/s * sin(35°) = 42.58 m/s
Using the formula for free fall motion, with initial velocity v_y and initial height h, we can find the time it takes for the shell to hit the ground:
h = v_y * t - (1/2) * g * t^2
where g is the acceleration due to gravity, approximately 9.8 m/s^2.
Rearranging the equation, we get:
0 = -4.9 t^2 + 42.58 t - 30
Solving this quadratic equation, we find that t ≈ 1.78 s.
Now, we can calculate the horizontal range by multiplying the horizontal component of the initial velocity by the time of flight:
range = v_x * t = 61.22 m/s * 1.78 s ≈ 109.02 m
Therefore, the horizontal range of the shell is approximately 109.02 m.
(b) To determine the velocity of the shell as it strikes the ground, we can use the equation for final velocity in free fall motion:
v_f = v_y - g * t
where v_f is the final velocity of the shell. Plugging in the given values, we have:
v_f = 42.58 m/s - 9.8 m/s^2 * 1.78 s
v_f ≈ 25.88 m/s
Therefore, the velocity of the shell as it strikes the ground is approximately 25.88 m/s.
A shell is fired from a cliff that is 30 m above a horizontal plane. The muzzle speed of the shell is 75.0 m/s and it
is fired at an elevation of 35 above the horizontal.
(a) Determine the horizontal range of the shell.
(b) Determine the velocity of the shell as it strikes the ground.
1 answer
1 answer