A ball is shot from a compressed-air gun at twice its terminalspeed. (For the following, assume the drag force depends on v2)
1. What is the magnitude of the ball’s initial acceleration, as amultiple of , if it is shotstraight up?Express your answer using one significantfigure.
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2. What is the magnitude of the ball’s initial acceleration, as amultiple of , if it is shotstraight down? Express your answer using one significantfigure.
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Scientists studying physics use a particular formula to calculate acceleration. The formula used to calculate acceleration is: Acceleration = change in speed or time
The drag force magnitude is determined by Fd = -1/2*rho*v2*Cd*A.
Where rho represents the air density, v the velocity, and Cd the drag coefficient.
A is the cross-sectional area. The terminal velocity, vt will be equal to the gravitational force.
1/2*rho*vt^2*Cd*A = m*g
Because the drag force is proportional with the square of velocity, doubling the velocity will increase the drag force by a factor of 4.
Fd = 1/2*rho*(2*vt)^2*Cd*A
Fd = 1/2*rho*4*vt2*Cd*A
Fd = 4*(1/2*rho*vt^2*Cd*A)
Fd = 4*m*g
The drag force is therefore 4 times greater than the acceleration caused by gravity. m*a = 4*m*g
a = 4*g
The acceleration is 4*g. The drag force will always oppose the motion of the ball. If the ball is shot straight up gravity will oppose the drag force. This is how it works.
a = 4*g + 5g
a = 5*g
The drag force acts in the opposite direction of gravity when the ball is shot downwards.
a = 4*g – g
a = 3*g
Number 1 = 5, and number 2 = 3.