# A baseball is hit at an initial speed of 40 m/s at an angle of 60° above the horizontal. The ball flies for 10 s. Find the distance, in meters, that the ball travels. Use g = 9.81 m/s2. A) 100 m B) 146 m C) 200 m D) 346 m

c) x = 200 m

Explanation:

Ball is projected at speed of 40 m/s at an angle of 60 degree

so here we have two components of velocity given as

now the horizontal distance moved by the ball in next 10 s is given as

so we have

Solution :

initial velocity=u=40 m/s

a=g=9.8m/s2

angle of projection=θ=60°

A) maximum Height h= u²sin²θ/2g

h= 40x40 sin60²/2x10

h= 40x40 x3/4x2 x 10

h=60m

B) Horizontal Range = X=u²sin2θ/g

=X= 40x40 sin2x60/10

=40x40 x√3/2x10

=80√3 m

## Related Questions

The sound intensity 50m from a wailing tornado siren is 0.1W/m2 . What is the sound intensity level 300m from the siren?

Explanation:

Looking at the formula of intensity we can see that it is inversely proportional to the square of distance and the other terms are all constant

So

We can now write the following equality

Substitute the know intensity at 50m we get

Consider light energy that is momentarily absorbed in glass and then re-emitted. Compared to the absorbed light, the frequency of the re-emitted light is A) considerably less.
B) slightly less.
C) the same.
D) slightly more.
E) considerably more.

Compared to the absorbed light, the frequency of the re-emitted light is the same.

Explanation:

The reason to why the light energy that is momentarily absorbed in glass and then re-emitted, compared to the absorbed light, the frequency of the re-emitted light is the same is because the wavelength of the absorbed light and the re-emitted light is the same. The energy carried by an electromagnetic wave is proportional to the frequency of the wave. The wavelength and frequency of the wave are connected via the speed of light:

c = fλ

where

• c is the speed of light
• f is the frequency of light
• λ is the wavelength of the light

Though light does not have mass, it does have energy and it's energy is conserved. Therefore, in the conservation of energy, the energy of the absorbed light must be equal to the re-emitted light. This is why the frequency of the re-emitted light is the same as the frequency of the absorbed light.

The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -50 °C and 900 Pa, respectively. (a)Determine the density of the Martian atmosphere for these conditions if the gas constant for the Martian atmosphere is assumed to be equivalent to that of carbon dioxide. (b) Compare the answer from part (a) with the density of the Earth’s atmosphere during a spring day when the temperature is 18 °C and the pressure 101.6 kPa

T = 273 + (-50) = 273 – 50 = 223 K

R = 188.82 J / kg K for CO2

Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg /

T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa

Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg /

(a) The density of the Martian atmosphere is 0.021kg/m^3

(b) The density of the Martian atmosphere (0.021kg/m^3) is less than the density of the Earth's atmosphere (1.217kg/m^3)

Explanation:

(a) Density of Martian atmosphere (D) = P/RT

P = 900 Pa, R = 189 J/kgK, T = -50°C = -50+273 = 223K

D = 900/189×223 = 900/42,147 = 0.021kg/m^3

b) Density of Earth's atmosphere (D) = P/RT

P = 101.6kPa = 101.6×1000 = 101,600 Pa, R = 287 J/kgK, T = 18°C = 18+273 = 291K

D = 101,600/287×291 = 101,600/83,517 = 1.217kg/m^3

The density of Martian atmosphere is less than the density of the Earth's atmosphere

According to missouri law, during what hours is it legal for a boat to tow a water-skier?