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masya89 [10]
3 years ago
12

During a baseball game, a batter hits a high pop-up. if the ball remains in the air for a total of 6.0 s, how high does it rise?

( Assume Accel. = 10 m/s^2 )
Physics
1 answer:
andrew11 [14]3 years ago
5 0
  • Time=6s
  • Vertex height's time= 6/2=3s
  • Acceleration due to gravity=10m/s^2

Apply second equation of kinematics

\\ \rm\hookrightarrow s=ut+\dfrac{1}{2}gt^2

\\ \rm\hookrightarrow s=1/2(10)(3)^2

\\ \rm\hookrightarrow s=5(9)

\\ \rm\hookrightarrow s=45m

It rises 45m

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high school physics, no need detail explain, just give the answer, but you have to make sure thank you
andrey2020 [161]

Answer:

approximately 30 degrees

Explanation:

If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component (v_y) of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):

v_f=v_i-a\,t\\v_f=v_y-g\,t\\0=v_y-9.8\,*\,2\\v_y=9.8\,*\,2=19.6 \frac{m}{s}

By knowing the vertical component of the initial velocity (19.6 m/s), and the actual magnitude of the total initial velocity (40 m/s), we can calculate what angle was the initial velocity vector forming above the horizontal. We use for such the fact that the sine of the angle relates the opposite side of a right angle triangle with the hypotenuse, and solve for the angle using the arcsin function:

sin(\theta)=\frac{opp}{hyp} \\sin(\theta)=\frac{19.6}{40}\\\theta=arcsin(\frac{19.6}{40})\\\theta=29.34^o

which tells us that the closer answer shown is 30^o

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4 years ago
Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vert
chubhunter [2.5K]

Answer:

362.41 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.8

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.04 m²

v = Terminal velocity

F = ma

F=\frac{1}{2}\rho CAv^2\\\Rightarrow mg=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{mg}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{20\times 9.81}{1.21\times 0.8\times 0.04}}\\\Rightarrow v=100.66924\ m/s

Converting to km/h

100.66924\times 3.6=362.41\ km/h

The terminal velocity of the stone is 362.41 km/h

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