# A stone thrown horizontally from the top of a 18-m tower hits the ground at a point 18 m from the base of the tower. (Ignore any effects due to air resistance.)

(a) Find the speed at which the stone was thrown.

m/s

(b) Find the speed of the stone just before it hits the ground.

m/s

## a. h = 0.5g*T^2

h = 18 m.

g = 9.8 m/s^2

T = Fall time in s.

Solve for T.

d = Xo*T

d = 18 m.

Xo = hor. speed in m/s.

T = Time calculated in part a.

Solve for Xo.

b. Y^2 = Yo + 2g*h

Y = Ver. component of Speed just before it hits the gnd.

Yo = 0 = Ver. component of initial speed.

g = 9.8 m/s^2

h = 18 m.

Solve for Y.

V = Sqrt(Xo^2 + Y^2) = Velocity just before it hits the gnd.

## (a) Well, let's see. The stone was thrown horizontally from the top of an 18-meter tower, and it hits the ground at a point 18 meters from the base of the tower. Now, if we ignore any effects due to air resistance, we can assume that the stone will travel in a straight line horizontally until it hits the ground. So, the time it takes for the stone to hit the ground is the same as the time it takes for the stone to travel horizontally 18 meters.

We can use the equation d = vt, where d is the distance, v is the velocity, and t is the time. In this case, d is 18 meters and v is what we are trying to find. We can assume that the time taken is the same, so we have:

18 = v * t

Since the stone was thrown horizontally, there is no vertical acceleration, and the only horizontal force is gravity. So, the time taken for the stone to hit the ground is the same as the time it would take for an object to fall 18 meters vertically.

Using the formula h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity, and t is the time, we can solve for time. Plugging in the values, we have:

18 = (1/2)(9.8)t^2

Multiplying both sides by 2 and dividing by 9.8, we get:

36/9.8 = t^2

t^2 = 36/9.8

Taking the square root of both sides, we have:

t = √(36/9.8)

Plugging this value of t back into the first equation, we get:

18 = v * √(36/9.8)

Now, solving for v:

v = 18 / √(36/9.8)

Calculating this, we find:

v ≈ 12.26 m/s

So, the speed at which the stone was thrown is approximately 12.26 m/s.

(b) Now, to find the speed of the stone just before it hits the ground, we can use the same equation, d = vt, but this time the distance is 18 meters (the distance from the base of the tower to the point where the stone hits the ground) and the time is the same as before. Plugging in these values, we have:

18 = v * √(36/9.8)

Now solving for v:

v = 18 / √(36/9.8)

Calculating this, we find:

v ≈ 12.26 m/s

So, the speed of the stone just before it hits the ground is also approximately 12.26 m/s.

Looks like the stone maintained its speed all the way down! Talk about consistency!

## To find the speed at which the stone was thrown and the speed of the stone just before it hits the ground, we can use the equations of motion.

Given:

Initial height of the stone (h) = 18 m

Distance traveled horizontally (x) = 18 m

(a) Find the speed at which the stone was thrown:

We can use the equation for the time of flight for a horizontally thrown projectile:

Time of flight (T) = x / (initial horizontal velocity) ----(1)

Since the stone is thrown horizontally, the initial vertical velocity is zero. Therefore, the time of flight is equal to the time taken to fall from the initial height (h) to the ground.

Using the equation for free fall:

h = (1/2) * g * t^2 ----(2)

where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time.

Substituting equation (2) into equation (1):

T = sqrt((2h) / g) ----(3)

Now, we can rearrange equation (1) to find the initial horizontal velocity (u):

Initial horizontal velocity (u) = x / T ----(4)

Substituting the given values into equation (3):

T = sqrt((2 * 18) / 9.8)

T = sqrt(3.673)

Substituting the values of T and x into equation (4):

Initial horizontal velocity (u) = 18 / sqrt(3.673)

Simplifying this equation will give us the speed at which the stone was thrown.

(b) Find the speed of the stone just before it hits the ground:

Since there is no horizontal force acting on the stone, the horizontal velocity remains constant throughout the motion. Therefore, the speed of the stone just before it hits the ground is equal to the initial horizontal velocity (u) calculated in part (a).

The speed of the stone just before it hits the ground is the same as the speed at which it was thrown.

Therefore, the speed of the stone just before it hits the ground is also 18 / sqrt(3.673) m/s.

## To solve this problem, we need to use the principles of projectile motion. In this case, we have a horizontally launched projectile, which means the initial vertical velocity is zero.

Let's break down the problem into two parts:

Part (a): Find the speed at which the stone was thrown.

We know that the stone was thrown horizontally, which means it only has initial horizontal velocity (Vx) and no initial vertical velocity. Since the vertical velocity is zero, we can use the following equation to find the time of flight (t):

h = (1/2) * g * t^2

Here, h is the vertical displacement (18 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.

Plugging in h = 18 m and g = -9.8 m/s^2 (negative because the stone is falling downwards), we can solve for t:

18 = (1/2) * (-9.8) * t^2

Simplifying the equation, we get:

t^2 = -36.72

Since time cannot be negative, we ignore the negative solution. Taking the square root, we find:

t = √36.72 ≈ 6.06 s

Now that we know the time of flight, we can use it to find the initial horizontal velocity (Vx). We can use the following equation:

Vx = dx / t

Here, dx is the horizontal displacement (18 m) and t is the time of flight (6.06 s). Plugging in the values, we get:

Vx = 18 / 6.06 ≈ 2.97 m/s

Therefore, the stone was thrown with a speed of approximately 2.97 m/s horizontally.

Part (b): Find the speed of the stone just before it hits the ground.

Since the vertical velocity increases due to the acceleration of gravity, we can use the following equation to find the final vertical velocity (Vy) just before hitting the ground:

Vy = g * t

Plugging in g = -9.8 m/s^2 and t = 6.06 s, we get:

Vy = -9.8 * 6.06 ≈ -59.55 m/s

Now, we can use the Pythagorean theorem to find the magnitude of the final velocity (Vf):

Vf = √(Vx^2 + Vy^2)

Plugging in Vx = 2.97 m/s and Vy = -59.55 m/s, we get:

Vf = √(2.97^2 + (-59.55)^2) ≈ 59.56 m/s

Therefore, the speed of the stone just before it hits the ground is approximately 59.56 m/s.