Question

A cannonball shell is launched with a velocity of 100 m/s at an angle of 30.0° above horizontal from a point on a cliff 50.0 m above a level plain below. How far from the base of the cliff does the shell strike the ground? There is no appreciable air resistance, and g = 9.80 m/s2 at the location of the cliff.

Answer 1

Answer:To find the horizontal distance traveled by the cannonball shell, we can use the equations of projectile motion.

Given:

Initial velocity (v₀) = 100 m/s

Launch angle (θ) = 30.0°

Height of the cliff (h) = 50.0 m

Acceleration due to gravity (g) = 9.80 m/s²

Step 1: Resolve the initial velocity into horizontal and vertical components.

The horizontal component of velocity (v₀x) can be found using v₀x = v₀ * cos(θ).

The vertical component of velocity (v₀y) can be found using v₀y = v₀ * sin(θ).

Step 2: Calculate the time taken to reach the ground.

The time of flight (t) can be found using the formula t = (2 * v₀y) / g.

Step 3: Find the horizontal distance traveled.

The horizontal distance (d) can be found using the formula d = v₀x * t.

Let's calculate the values step-by-step:

Step 1:

v₀x = 100 m/s * cos(30.0°) = 100 m/s * 0.866 = 86.6 m/s (rounded to one decimal place)

v₀y = 100 m/s * sin(30.0°) = 100 m/s * 0.5 = 50.0 m/s

Step 2:

t = (2 * v₀y) / g = (2 * 50.0 m/s) / 9.80 m/s² = 10.2 s (rounded to one decimal place)

Step 3:

d = v₀x * t = 86.6 m/s * 10.2 s = 883.3 m (rounded to one decimal place)

Therefore, the cannonball shell strikes the ground approximately 883.3 meters away from the base of the cliff.

Step-by-step explanation: