Question

A rock is thrown from the edge of the top of a 100-ft tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 160 ft from the base of the building 5.0 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown

Answer 1

To find the initial speed at which the rock was thrown from the building, calculate the horizontal and vertical components of motion separately, then use the Pythagorean theorem to combine them. The rock was thrown with an initial speed of approximately 32.25 ft/s.

Finding the Initial Speed of the Thrown Rock

To calculate the initial speed with which the rock was thrown, we need to utilize the known information about the motion and apply kinematic equations. The rock is thrown from the edge of a 100-ft tall building and lands 160 ft horizontally from the base after 5.0 s. Keeping air resistance out of the question, the motion of the rock can be divided into two components: horizontal (constant velocity) and vertical (accelerated motion).

First, to find the horizontal speed (ux), we use the formula for constant velocity:

• Distance = Speed × Time

ux = 160 ft / 5.0 s = 32 ft/s.

For the vertical speed (uy), we apply the kinematic equation for distance under constant acceleration:

• s = uyt + (1/2)at2

Here, s is the displacement (100 ft, but it's downward so we take it as negative), uy is the initial vertical speed (which we're solving for), a is the acceleration due to gravity (approximately 32 ft/s2), and t is the time (5.0 s).

By inserting the known values and solving for uy, we find:

-100 ft = uy(5.0 s) + (1/2)(-32 ft/s2)(5.0 s)2

uy = -4 ft/s (upward).

Finally, to determine the initial speed of the throw, we combine the horizontal and vertical components with the Pythagorean theorem:

• Initial Speed = √(ux2 + uy2)
  

Initial Speed = √((32 ft/s)2 + (-4 ft/s)2) = √(1024 + 16) ft/s

Initial Speed = √(1040) ft/s ≈ 32.25 ft/s

The rock was thrown with an initial speed of approximately 32.25 ft/s.

Answer 2

68ft

Step-by-step explanation:

x-direction

Δx = vt

160 = v(5)

v = 32

y-direction

Δx =
`v_(0y) t + (1)/(2) a_(y) t^(2)`

*note that when using ft, acceleration by gravity is -32, not -9.8

-100 = v(5) + 0.5(-32)(5^2)

v = 60

adding horizontal & vertical vectors to find resultant

use pythagreon theorem

32^2 + 60^2 = h^2

h = 68