Aslingshot fires a pebble from the top of a building at a speed of 14.7 m/s. the building is 36.0 m tall. ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired in the following directions. (a) horizontally m/s (b) vertically straight up m/s (c) vertically straight down m/s

The final speed is 30.37 m/s for the three directions.

Explanation:

Given the initial speed, v₀ = 14.7 m/s and the height of the building as 36.0 m (neglecting air resistance and height of slingshot), we apply the principle of Conversation of Mechanical Energy:

E = Kinetic Energy (k.E) + Potential Energy (P.E)

K.E = 0.5mv²

P.E = mgh

∴ 0.5mv₀² + mgh₀ = 0.5mv₁² + mgh₁

Since the final height (h₁) is zero and m is a common term, then we can re-write the above equation as:

0.5v₀² + gh₀ = 0.5v₁² + 0

0.5v₁² - 0.5v₀² = gh₀

0.5(v₁² - v₀²) = gh₀

v₁² = v₀² + 2gh₀

Fitting in the given terms, we can calculate the final speed

v₁² = (14.7)² + (2)(9.81)(36.0)

   = 216.09 + 706.32 = 922.41

v₁ = √922.41 = 30.37 m/s.

The speed at which the pebble hits the ground remains the same because kinetic energy depends on the speed of the object and not the direction to which the object is thrown.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s