A stone is thrown from the top of a building upward at an angle of 30.0∘to the horizontal and with an initial speed of 20.0 m/s. If the height of the building is 45.0 m, what is the speed in m/sec of the stone just before it strikes the ground ?

u = 20 cos 30 until it hits the ground.

Vi = initial vertical velocity = +20 sin 30 = +10

initial vertical displacement = 45
final vertical displacement = 0

0 = 45 + Vi t - 4.9 t^2
solve for t

v = Vi - 9.8 t
solve for v

speed = sqrt (u^2 + v^2)

Well, it seems like the stone wants to make a grand entrance!

To solve this question, we can break it down into two components: horizontal and vertical.

First, let's find the time it takes for the stone to reach the ground. We can use the vertical motion equation:

h = v₀y * t - (1/2) * g * t²

where h is the height (45 m), v₀y is the initial vertical velocity (20 m/s * sin(30°)), g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time we're looking for.

Now, let's find the final horizontal velocity using the horizontal component:

v₀x = v₀ * cos(30°)

where v₀ is the initial velocity (20 m/s).

Since there is no horizontal acceleration (ignoring air resistance), the horizontal velocity will remain constant throughout. So, the final horizontal velocity will also be v₀x.

Finally, we can find the final speed just before the stone strikes the ground using the Pythagorean theorem:

v = √(v₀x² + v₀y²)

Armed with these calculations, the stone's speed just before it strikes the ground is no joke: approximately 17.4 m/s!

To find the speed of the stone just before it strikes the ground, we can break down the problem into two components: the horizontal component and the vertical component.

Step 1: Find the time it takes for the stone to reach the ground.
To find the time, we can use the vertical motion equation:

h = ut + (1/2)gt^2

Where:
h = height of the building = 45.0 m
u = initial vertical velocity (for upward motion) = 20.0 m/s
g = acceleration due to gravity = 9.8 m/s^2
t = time

Rearranging the equation:

45.0 = (20.0)t + (1/2)(9.8)(t^2)

This is a quadratic equation in terms of t. Solving for t, we get:

(4.9)t^2 + (20.0)t - 45.0 = 0

Using the quadratic formula, we find two solutions for t. Since we are looking for the time it takes for the stone to reach the ground, we take the positive root:

t = (-20.0 + √(20.0^2 - 4 * 4.9 * (-45.0))) / (2 * 4.9)

t ≈ 2.92 seconds

Step 2: Find the horizontal distance covered by the stone.
To find the horizontal distance covered (range), we can use the horizontal motion equation:

Range = horizontal velocity × time

The horizontal velocity remains constant throughout the motion and is given by:

vh = initial velocity (for horizontal motion) × cos(angle)

Where:
initial velocity (for horizontal motion) = 20.0 m/s (the same as the initial vertical velocity)
angle = 30.0 degrees

vh = 20.0 m/s × cos(30.0)
vh ≈ 17.32 m/s

Now we can find the horizontal distance covered, given by:

Range = vh × t

Range = 17.32 m/s × 2.92 s
Range ≈ 50.55 meters

Therefore, just before striking the ground, the stone is approximately 50.55 meters horizontally away from the building's base.

Step 3: Find the speed just before the stone strikes the ground.
To find the speed, we can apply the Pythagorean theorem. The speed will be the magnitude of the total velocity just before the stone strikes the ground.

total velocity = √(horizontal velocity^2 + vertical velocity^2)

For the horizontal velocity, we already calculated it as 17.32 m/s.
For the vertical velocity, we can use another vertical motion equation:

vertical velocity = initial velocity (for vertical motion) - g × t

Where:
initial velocity (for vertical motion) = 20.0 m/s
g = acceleration due to gravity = 9.8 m/s^2
t = time = 2.92 s

vertical velocity = 20.0 m/s - 9.8 m/s^2 × 2.92 s
vertical velocity ≈ -12.18 m/s (negative because it is pointing downward)

Now we can find the speed:

speed = √(17.32^2 + (-12.18)^2)
speed ≈ 21.80 m/s

Therefore, the speed of the stone just before it strikes the ground is approximately 21.80 m/s.

To find the speed of the stone just before it strikes the ground, you can use the principle of conservation of energy. The initial potential energy of the stone at the top of the building is equal to the sum of its final kinetic energy just before it hits the ground and the potential energy at the ground level.

Let's break down the problem into steps:

Step 1: Find the initial vertical velocity component (Vy) of the stone.
Given that the stone is thrown at an angle of 30.0∘ to the horizontal and with an initial speed of 20.0 m/s, we can apply trigonometry to find the vertical component of the initial velocity.
Vy = V * sin(θ), where V is the initial speed and θ is the angle.
Vy = 20.0 m/s * sin(30.0∘) = 10.0 m/s.

Step 2: Find the time taken by the stone to reach the ground.
We can use the kinematic equation for vertical motion: h = Vyo * t + (1/2) * g * t^2, where h is the height, Vyo is the initial vertical velocity component, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Rearranging the equation, we get t^2 + (Vyo/g) * t - (2h/g) = 0.
Solving this quadratic equation gives us t = 3.0 seconds (ignoring the negative value, as time cannot be negative).

Step 3: Find the initial horizontal velocity component (Vx) of the stone.
Given that the stone is thrown at an angle of 30.0∘ to the horizontal and with an initial speed of 20.0 m/s, we can again apply trigonometry to find the horizontal component of the initial velocity.
Vx = V * cos(θ), where V is the initial speed and θ is the angle.
Vx = 20.0 m/s * cos(30.0∘) = 17.32 m/s.

Step 4: Find the horizontal distance traveled by the stone.
Using the time calculated in Step 2 and the initial horizontal velocity component (Vx), we can calculate the horizontal distance (R) traveled by the stone.
R = Vx * t = 17.32 m/s * 3.0 s = 51.96 m.

Step 5: Find the final vertical velocity component (Vf) just before hitting the ground.
Using the kinematic equation for vertical motion: Vf = Vyo + g * t, where Vyo is the initial vertical velocity component and g is the acceleration due to gravity.
Vf = 10.0 m/s + (9.8 m/s^2 * 3.0 s) = 39.4 m/s.

Step 6: Find the final speed of the stone just before it strikes the ground.
To find the final speed (Vf) of the stone just before it strikes the ground, we need to calculate the resultant velocity of the vertical and horizontal components.
Vf = √(Vx^2 + Vf^2)
Vf = √(17.32 m/s)^2 + (39.4 m/s)^2) = √(299.9224 + 1552.3600) = √1852.2824 = 43.0 m/s (approximately).

Therefore, the speed of the stone just before it strikes the ground is approximately 43.0 m/s.