Web Analytics
Answered

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?

Verified Answer :

Hippalectryon

(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 \(t=\frac{v_0\sin(\alpha)+\sqrt{(v_0\sin(\alpha))^2+gd}}g\), thus the speed at that time is \(v=-\sqrt{(v_0\sin(\alpha))^2+gd}\), 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.

a) \(v_f=34m/s\)

b)The impact velocity will be the same.

Given:

d=8.0m

\(v_0\)=22m/s

Angle= \(31^o\)

a)

Let \(y(t)\) be the stone's height in function of time, then by Newton's second law:

\(y(t)=d-v_0sin(\alpha)-\frac{1}{2}gt^2\)

where

\(\alpha=31^0\)  is the throw's angle

When stone hits the ground then \(y(t)=0\) which happened when:

\(t=\frac{v_0sin(\alpha)+\sqrt{v_0sin(\alpha)^2+gd}}{g}\)

Therefore, the speed at that time is;

\(v=-\sqrt{v_0sin(\alpha)^2+gd}\)

So, the value of \(v_f=34m/s\).

b) Impact velocity

The impact velocity will be the same.

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.

Find more information about "Impact velocity" here:

brainly.com/question/4931057

Related Questions