To solve this problem, we can use the equations of motion for projectile motion. The key variables we need to find are the horizontal range (x) and the velocity of the shell as it strikes the ground (v).
(a) To determine the horizontal range, we need to find the time of flight (t) first. We can use the vertical motion equation:
y = yo + voyt - 1/2gt^2
Where y = 0 (since the shell strikes the ground), yo = 30 m, voy = v*sin(theta), g = 9.8 m/s^2, and theta = 35°.
0 = 30 + (v*sin(35°))t - 1/2(9.8)t^2
Simplifying, we get:
4.9t^2 - (v*sin(35°))t - 30 = 0
This is a quadratic equation in terms of t. Solving this equation will give us two possible values for t. We can discard the negative value since time cannot be negative.
Using the quadratic formula, we get:
t = (-(v*sin(35°)) + sqrt((v*sin(35°))^2 - 4(4.9)(-30))) / (2(4.9))
Simplifying, we get:
t ≈ 5.57 s
Now that we have the time of flight, we can find the horizontal range using the equation:
x = voxt
Where x is the horizontal range, vox = v*cos(theta), and t = 5.57 s.
x = (v*cos(35°)) * 5.57
Simplifying, we get:
x ≈ 387.6 m
Therefore, the horizontal range of the shell is approximately 387.6 m.
(b) To determine the velocity of the shell as it strikes the ground, we can use the x and y components of velocity. The x component remains constant throughout the motion, while the y component changes due to acceleration.
The x component of velocity (vox) is given by:
vox = v*cos(theta)
Plugging in the values, we get:
vox = 75*cos(35°)
Simplifying, we get:
vox ≈ 61.3 m/s
The y component of velocity (voy) can be found using:
voy = v*sin(theta)
Plugging in the values, we get:
voy = 75*sin(35°)
Simplifying, we get:
voy ≈ 42.7 m/s
At the point of impact, the y component of velocity (vfy) is equal to -voy (since the velocity is downward).
Using the equation:
vfy = voy - gt
We can solve for vfy:
vfy = -voy + g*t
Plugging in the values, we get:
vfy = -42.7 + (9.8)(5.57)
Simplifying, we get:
vfy ≈ -45.7 m/s
The magnitude of the velocity vector v is given by the Pythagorean theorem:
v = sqrt(vox^2 + vfy^2)
Plugging in the values, we get:
v = sqrt((61.3)^2 + (-45.7)^2)
Using a calculator, we get:
v ≈ 76.3 m/s
Therefore, the velocity of the shell as it strikes the ground is approximately 76.3 m/s.
Assignment 4 Projectile Motions
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