Question

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.

Answer 1

To solve this problem, we need to split the initial velocity into its horizontal and vertical components.

Given:
Initial vertical velocity (Vy) = 75.0 m/s * sin(35°)
Initial horizontal velocity (Vx) = 75.0 m/s * cos(35°)

(a) To find the horizontal range, we need to find the time of flight (t) first. The time of flight can be calculated using the following formula:

t = (2 * Vy) / g

where g is the acceleration due to gravity (9.8 m/s^2).

t = (2 * 75.0 m/s * sin(35°)) / 9.8 m/s^2
t ≈ 11.65 s

Now, we can calculate the horizontal range (R) using the formula:

R = Vx * t

R = (75.0 m/s * cos(35°)) * 11.65 s
R ≈ 713.6 m

Therefore, the horizontal range of the shell is approximately 713.6 meters.

(b) To find the velocity of the shell as it strikes the ground, we can use the following formula:

V = √(Vx^2 + Vy^2)

V = √((75.0 m/s * cos(35°))^2 + (75.0 m/s * sin(35°))^2)
V ≈ 75.0 m/s

Therefore, the velocity of the shell as it strikes the ground is approximately 75.0 m/s.