A slingshot fires a pebble from the top of a building at a speed of 16.0 m/s. The building is 32.0 m tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

To find the speed at which the pebble strikes the ground in each case, we can use the principles of projectile motion.

(a) When the pebble is fired horizontally:
In this case, the initial velocity of the pebble in the vertical direction is zero, as there is no upward or downward component of the velocity. However, the horizontal component of the velocity remains constant.

Using the equation of motion for vertical motion, we can find the time it takes for the pebble to fall from the top of the building to the ground:
h = (1/2) * g * t^2, where h is the height of the building (32.0 m) and g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation, we get:
t = sqrt((2 * h) / g)

Substituting the values, we get:
t = sqrt((2 * 32.0) / 9.8) ≈ 2.02 s

Since the horizontal velocity remains constant, we can find the horizontal distance covered by multiplying the horizontal velocity (16.0 m/s) by the time:
d = v * t: d = 16.0 * 2.02 ≈ 32.3 m

Therefore, the pebble strikes the ground with a speed of approximately 32.3 m/s when fired horizontally.

(b) When the pebble is fired vertically straight up:
In this case, the horizontal component of the velocity is zero, and only the vertical component matters. Initially, the pebble starts with a velocity of 16.0 m/s in the upward direction.

Using the equation of motion for vertical motion, we can find the final velocity when the pebble reaches the ground:
v_f^2 = v_i^2 + 2 * g * h, where v_i is the initial velocity (16.0 m/s), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the building (32.0 m).
Substituting the values, we get:
v_f^2 = (16.0)^2 + 2 * 9.8 * 32.0
v_f^2 ≈ 384.4
v_f ≈ sqrt(384.4) ≈ 19.6 m/s

Therefore, the pebble strikes the ground with a speed of approximately 19.6 m/s when fired vertically straight up.

(c) When the pebble is fired vertically straight down:
Similar to the vertical straight up case, the horizontal component of the velocity is zero. However, the initial velocity is in the downward direction this time.

Using the same equation of motion for vertical motion, we can find the final velocity when the pebble reaches the ground:
v_f^2 = v_i^2 + 2 * g * h, where v_i is the initial velocity in the downward direction (16.0 m/s), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the building (32.0 m).
Substituting the values, we get:
v_f^2 = (-16.0)^2 + 2 * 9.8 * 32.0
v_f^2 ≈ 1216.4
v_f ≈ sqrt(1216.4) ≈ 34.9 m/s

Therefore, the pebble strikes the ground with a speed of approximately 34.9 m/s when fired vertically straight down.