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 can break the initial velocity of the shell into its horizontal and vertical components.

Given:
Initial vertical velocity (Vy) = 75.0 * sin(35°)
Initial horizontal velocity (Vx) = 75.0 * cos(35°)
Initial height (h) = 30.0 m

a) To determine the horizontal range of the shell, we need to find the time of flight (T). We can use the formula:
T = (2 * Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).

T = (2 * (75.0 * sin(35°))) / 9.8

To find the horizontal range (R), we use the formula:
R = Vx * T

Substituting the values:
R = (75.0 * cos(35°)) * [(2 * (75.0 * sin(35°))) / 9.8]
Solve for R to get the horizontal range of the shell.

b) To determine the velocity of the shell as it strikes the ground, we need to find the final velocity (Vf) in the y-direction when the shell reaches the ground.

Using the formula:
Vf = Vy + gt
where t is the time of flight (T).

Vf = (75.0 * sin(35°)) + (9.8 * T)

Substitute the value of T we obtained in part a) to find the final y-direction velocity.

Finally, to find the velocity of the shell as it strikes the ground, we use the Pythagorean theorem:
V = √(Vx² + Vf²)

Substitute the values of Vx and Vf to calculate the final velocity of the shell as it strikes the ground.