The image shows the trajectory of baseball, with four labelled positions A throu

Question

The image shows the trajectory of baseball, with four labelled positions A through D. Positions A and C are at the same height, position B is at the highest point of the trajectory, and position D is shortly before the player catches the ball. Neglect air resistance for this problem. Which one of the following statements is true? The vertical component of the acceleration changes sign at point B. The ball has zero acceleration at point B. The ball has zero velocity at point B. The speed of the ball at point A is the same as at point C. The vertical component of the velocity is the same at points A and C. Positions B and D are 11.9 meters and 1.9 meters above the playing field, respectively. If the ball had a speed of 6.5 m/s in position B, what is its speed in position D? (in m/s) A: 5.1 B: 6.3 C. 7.9 D: 9.9 E: 12.4 F: 15.4 G: 19.3 H: 24.1

Explanation / Answer

Yes at the maximum height the accelaration due to gravity changes the sign ,so the statementis true
and at maximum height,there is a Horizontal component of velocity

since A and C are at same height from ground so it has same speed at two points A and C
and also vertical component of velocity is same at points A and C exceot in sign

so true statements are

given numbers to the statements from top to bottom as 1,2 ,3,4 and 5

so the answer is 1,3 and 5 are true

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speed at B is Vo*cos(theta)

Vo is the initial velocity of the ball and theta is the angle between horizontal line and initial velocity

So Vo*cos(theta) = 6.5 ........(1)

B is at a height of Hmax = 11.9 m

But Hmax = Vo^2*sin^2(theta)/2g = 11.9

Vo*sin(theta) = sqrt(11.9*2*9.8) = 15.3 m/sec........(2)

(2) / (1)

tan(theta) = 15.3/6.5

theta = tan^(-1)(15.3/6.5) = 67 degrees

so vertical distance travelled before catching is 11.9 - 1.9 = 10 m

so using kinematic equations

V^2 - Vo^2 = 2*a*S

along y-axis

Vy^2 - 0^2 = 2*9.8*10

vy = 14 m/sec

Vx = Vox = 6.5 m/sec

so velocity at D is VD = sqrt(6.5^2+14^2) = 15.4 m/sec

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