Question

***PLEASE HELP!!!!!!!***

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 = 22 m/s at an angle= 31degrees below horizontal. The stone moves without air resistance; use a Cartesian coordinate system with the origin at the stone's initial position.

50% 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?

Answer 1

The impact speed vf can be calculated using kinematic equations, and for part (b), the impact velocity would be the same if thrown at the same angle above the horizontal with the same initial speed.

Step-by-step explanation:

To determine the speed vf at which the stone strikes the ground thrown from a vertical height d = 8.0 m with an initial velocity v0 = 22 m/s at a 31-degree angle below the horizontal, we can use kinematic equations for projectile motion. Since there is no air resistance, the horizontal and vertical components of the motion can be treated separately.

First, resolve the initial velocity into horizontal (vx) and vertical (vy) components using trigonometry:

• vx = v0 × cos(θ) = 22 m/s × cos(31°)
• vy = v0 × sin(θ) = 22 m/s × sin(-31°) (negative since it is below the horizontal)

The final vertical velocity vfy can be found using the following kinematic equation, considering the downward direction as positive:

vfy2 = vy2 + 2gd

The horizontal velocity remains constant (vx) since there is no air resistance. Hence, the final horizontal velocity vfx = vx. The final speed is the magnitude of the resultant of vfx and vfy, which can be found using the Pythagorean theorem:

vf = sqrt(vfx2 + vfy2)

For part (b), due to symmetry in projectile motion, the impact velocity would be the same whether the stone is thrown at the same angle above or below the horizontal with the same initial speed. Therefore, if the stone had been thrown with the same initial speed and angle but above the horizontal, its impact velocity would not be different.

Answer 2

(a)

Let
`y(t)` be the stone's height in function of time, then by Newton's second law
`y(t)=d-v_0\sin(\alpha)-\frac12gt^2` where
`\alpha=31^\circ` is the throw's angle.

The stone hits the ground when
`y(t)=0` which occurs when
`t=\frac{v_0\sin(\alpha)+√((v_0\sin(\alpha))^2+gd)}g`, thus the speed at that time is
`v=-√((v_0\sin(\alpha))^2+gd)`, which gives
`\boxed{v_f\approx34m/s}` for
`g=10m/s^2`

(b)

The impact velocity will be the same.

Indeed, if you throw the rock upwards it will first decelerate, then reach a maximum height and then accelerate again. When it reaches the height
`d=8m` for the second time, it will have a vertical speed opposite to what it had at launch, and therefore the situation will be the same as if we had thrown the thrown downward from that new position.