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

Answer:

a) Vf = 27.13 m/s

b) It would have been the same

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:

$Vf=\sqrt{(-15.3)^2+(24*cos(21))^2}$

$Vf=27.13m/s$

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.