All categories

3 years ago

Help with physics projectile motion

Physics

1 answer:

BlackZzzverrR [31]3 years ago

3 0

Answer:

10.4 m/s

Explanation:

First, find the time it takes for the projectile to fall 6 m.

Given:

y₀ = 6 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.11 s

Now find the horizontal position of the target after that time:

Given:

x₀ = 6 m

v₀ = 5 m/s

a = 0 m/s²

t = 1.11 s

Find: x

x = x₀ + v₀ t + ½ at²

x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²

x = 11.5 m

Finally, find the launch velocity needed to travel that distance in that time.

Given:

x₀ = 0 m

x = 11.5 m

t = 1.11 s

a = 0 m/s²

Find: v₀

(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²

v₀ = 10.4 m/s

You might be interested in

Dry air will break down if the electric field exceeds about 3.0×106v/m. part a what amount of charge can be placed on a capacito

Genrish500 [490]

The solution for this problem is:The charge would be now equal to:(electric constant) multiplied by the (field strength) multiplied by the (area) so plugging in our values, will give us:8.85 * 10^-12 As / (V * m) * 3 * 10^6 V/m * 0.055 m^2 = 1.46 e-6 amperes would be the answer

7 0

3 years ago

Please help I’m so confused

Ilya [14]

1.wavelength

2.trough

3.amplitude

4.crest

5 0

3 years ago

Which of the following is a personal decision that involves science?

malfutka [58]

A because its the only one that actually involves science

4 0

2 years ago

Can a black light be used to charge a solar powered product

Goshia [24]

Yes It can because the black light would be considered solar powered

6 0

3 years ago

Two equally charged insulating balls each weigh 0.16 g and hang from a common point by identical threads 35 cm long. The balls r

Dafna11 [192]

Answer:

Q₁ = Q₂ = 8.84 x 10⁻⁹ C

Explanation:

given,

mass of ball, m = 0.16 g = 1.6 x 10⁻⁴ Kg

ball each other, r = 6.8 cm

Weight of the ball

F_w = m g

F_w = 1.6 x 10⁻⁴ x 9.8

F_w = 1.56 x 10⁻³ N

The tension in each string is a force directed along the length of the string and is the hypotenuse of a right triangle.

we have to find the horizontal component of the forces.

The length of the string,L is 35 cm so, it will be the hypotenuse.

θ be the angle made with imaginary vertical line and the string.

now,

$sin \theta = \dfrac{r\2}{L}$

$sin \theta = \dfrac{3.4}{35}$

θ = 5.57°

horizontal component of the force = ?

vertical component of force,F_v = 1.56 x 10⁻³ N

$tan\theta = \dfrac{F_H}{F_v}$

$tan(5.57^0) = \dfrac{F_H}{1.56\times 10^{-3}}$

F_h = 1.52 x 10⁻⁴ N

now, each ball will be repelled by

F = 1.52 x 10⁻⁴ N

now calculation of charges

$F = \dfrac{kQ_1Q_2}{r^2}$

Q₁ = Q₂ because both charge are same

$1.52\times 10^{-4} = \dfrac{9\times10^9Q^2}{0.068^2}$

Q² = 7.809 x 10⁻¹⁷

Q = 8.84 x 10⁻⁹ C

hence the change on the balls were Q₁ = Q₂ = 8.84 x 10⁻⁹ C

5 0

3 years ago

Help with physics projectile motion​

Help with physics projectile motion