A ball is falling after rolling off a tall roof. The ball has

2 answers:

8 0

C.
Because it’s falling it has acceleration in the y direction. If you have acceleration, you usually also have velocity, and since kinetic energy is KE= Mv^2 you know you have it. It also has potential energy because it has some height to it, and PE= Mgh.

Stella [2.4K]3 years ago

8 0

The ball has both kinetic and potential energy.

You might be interested in

Calcular la longitud del faldón de una Rampa de Acceso , que en planta tiene una longitud de 20 m y la pendiente es 27%.

seraphim [82]

La longitud del faldón de la rampa es de 5.4 m.

La pendiente expresada en porcentaje sigue la siguiente ecuación:

$m=\frac{y}{x}*100$ (1)

Donde:

y es la elevacion de la rampa (faldón)
x es la longitud de la ramapa (20 m)

Sabemos que la pendiente es de 27%. Por lo tanto, usando la ecuación 1, despejamos y.

$27=\frac{y}{20}*100$

$y=\frac{27*20}{100}$

$y=5.4\: m$

La longitud del faldón es 5.4 m

Pudes ver más sobre el tema aquí:

brainly.com/question/8906330

5 0

3 years ago

A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?

Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

$H_m = \frac{u^2sin^2 \theta}{2g}$

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

$H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta = sin^{-1}(0.2123)\\\\\theta = 12.26^0$