Question

A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 24 m/s at an angle θ = 21 ° below horizontal. The stone moves without air resistance; use a Cartesian coordinate system with the origin at the stone's initial position.

Part (a) With what speed, vf in meters per second, does the stone strike the ground 50%
Part (b) If the stone had been thrown from the clifftop with the same initial speed and the same angle, but above the horizontal would its impact velocity be different? YesNo Grade Summary 0% 100% Potential

Answer 1

To determine the speed at which the stone strikes the ground, we can break the initial velocity into its horizontal and vertical components and calculate the time and horizontal distance traveled by the stone. By using the equations for vertical and horizontal motion and the initial conditions provided, we can calculate the speed at which the stone strikes the ground. If the stone had been thrown from a greater height above the horizontal, the impact velocity would be different as the horizontal component of the velocity would be greater.

Step-by-step explanation:

To determine the speed at which the stone strikes the ground, we need to analyze the motion of the stone both horizontally and vertically. Since the stone is thrown with an initial velocity of 24 m/s at an angle of 21° below the horizontal, we can break the initial velocity into its horizontal and vertical components. The horizontal component, Vx, can be calculated using Vx = v0 * cos(θ), where v0 is the initial velocity and θ is the angle. The vertical component, Vy, can be calculated using Vy = v0 * sin(θ).

In this case, Vx = 24 m/s * cos(21°) = 22.4 m/s and Vy = 24 m/s * sin(21°) = 8.7 m/s. The time it takes for the stone to hit the ground can be calculated using the equation y = y0 + Vy0t + 0.5gt^2, where y0 is the initial vertical position, Vy0 is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. In this case, y0 = 8.0 m, Vy0 = 8.7 m/s, and g = 9.8 m/s^2. We can solve for t using the quadratic equation.

Once we have the time, we can calculate the horizontal distance traveled by the stone using x = Vx0t. Since Vx0 is constant and equal to Vx, we can substitute Vx into the equation. In this case, x = 22.4 m/s * t.

Therefore, the speed vf at which the stone strikes the ground can be calculated using vf = √(Vx^2 + Vy^2). We can substitute the values we calculated for Vx and Vy to find the answer.

For part (b) of the question, if the stone had been thrown from the clifftop with the same initial speed and the same angle but above the horizontal, the impact velocity will be different. The vertical component of the velocity will be the same, as it only depends on the initial speed and angle of projection. However, the horizontal component will be different. The horizontal component of the velocity depends on the height of the cliff. When thrown from a higher height, the horizontal component will be greater, resulting in a higher impact velocity.

Answer 2

a) Vf = 27.13 m/s

b) It would have been the same

Step-by-step explanation:

On the y-axis:

`Y=-Vo*sin\theta*t-1/2*g*t^2`

`-8=-24*sin(21)*t-1/2*10*t^2`

Solving for t:

t1 = 0.67s t2= -2.4s

Discarding the negative value and using the positive one to calculate the velocity:

`Vf_y = -Vo*sin\theta-g*t`

`Vf_y = -15.3m/s`

So, the module of the velocity will be:

`Vf=√((-15.3)^2+(24*cos(21))^2)`

`Vf=27.13m/s`

If you throw it above horizontal, it would go up first, and when it reached the initial height, the velocity would be the same at the throwing instant. And starting then, the movement will be the same.