Question

A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

Answer 1

Answer:

vf = 30 m/s : (the magnitude of the velocity of the stone just before it hits the ground)

Explanation:

Because the stone moves with uniformly accelerated movement we apply the following formulas:

vf²=v₀²+2*g*h Formula (1)

Where:  

h: displacement in meters (m)  

v₀: initial speed in m/s

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

Free fall of the stone

Data

v₀ =  10 m/s

vf =  30.0 m/s

g = 9,8 m/s²

We replace data in the formula (1) to calculate h:

vf²=v₀²+2*g*h

(30)² = (10)² + (2)(9.8)*h

(30)²- (10)²= (2)(9.8)*h

h =( (30)²- (10)²) /( 2)(9.8)

h = 40.816 m

Semiparabolic movement of the stone

Data

v₀x =  10 m/s

v₀y =  0 m/s

g = 9.8 m/s²

h= 40.816 m

We replace data in the formula (1) to calculate vfy :

vfy² = v₀y² + 2*g*h

vfy² = 0 + (2)(9.8)( 40.816)

$v_{fy}=\sqrt{2*9.8*40.816} = 28.284 \frac{m}{s}$

$v_{f}=\sqrt{v_{ox}^2+v_{fy}^2}=\sqrt{(10)^2+(28.284)^2} = 30\frac{m}{s}$