# Help with physics questions please?

1.  At the instant a bullet (Bullet A) is being fired horizontally from a gun, an identical bullet (Bullet B) is dropped from the same height as the muzzle of the firing gun. Briefly describe the order in which bullets A and B hit the ground.

5. A softball coach throws a 1 kg ball horizontally to her player with an initial velocity of 40 m/s. The ball leaves her hand at 6 m high, and her catcher is 20 meters away.

A. What is the vertical acceleration, ay, in m/s2? (+ is for up, - is for down)

B. How long is the ball in the air?

10. An M-16 is fired from level ground at an upwards angle of 10 degrees with respect to the horizontal. The initial velocity of the bullet is 948 m/s, but because it is fired at a 10 degree angle, the vertical component of the initial velocity is 165 m/s and the horizontal component is 934 m/s. Ignore air resistance and the curvature of the earth in this problem.

A. What is the initial vertical velocity, vyi, in m/s?  (+ is for up, - is for down)

B. What is the vertical acceleration, ay, in m/s2? (+ is for up, - is for down)

C. What is the final vertical velocity, vxf, in m/s?

D. What is the height difference, Δy, between the point where the bullet is fired and the point where the bullet lands?

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1. At the instant a bullet  A is being fired horizontally from a gun, an identical bullet (Bullet B) is dropped from the same height as the muzzle of the firing gun. Briefly describe the order in which bullets A and B hit the ground.

They heat the ground simultaneously, since they have the same initial vertical speed 5. A softball coach throws a 1 kg ball horizontally to her player with an initial velocity of 40 m/s. The ball leaves her hand at 6 m high, and her catcher is 20 meters away.A. What is the vertical acceleration, ay, in m/s2? (+ is for up, - is for down)ay = -9.807 m/sec^2B. How long (t) is the ball in the air?t = d /Vox = 20/40 = 0.5 seccatching height h' = 6-g/2*t^2 = 6-4.903*0.25 = 4.77 m if the catcher miss the ball , flying time t' will last √ 2h/g = √ 12/9.807 = 1.106 sec 10. An M-16 is fired from level ground at an upwards angle of 10 degrees with respect to the horizontal. The initial velocity of the bullet is 948 m/s, but because it is fired at a 10 degree angle, the vertical component of the initial velocity is 165 m/s and the horizontal component is 934 m/s. Ignore air resistance and the curvature of the earth in this problem. A. What is the initial vertical velocity, vyi, in m/s? (+ is for up, - is for down)Vyi = Vi*sin 10° = 164.6 m/sec (to be precise)B. What is the vertical acceleration, ay, in m/s2? (+ is for up, - is for down)ay = g = -9.807 m/sec^2C. What is the final vertical velocity, vxf, in m/s?Vxf = Vxi = 948*cos 10° = 933.6 m/sec (to be precise)D. What is the height difference, Δy, between the point where the bullet is fired and the point where the bullet lands?

No difference , if not stated

flying time t = 2*Vyi/g = 2*164.6 / 9.807 = 33.57 sec

theoretical range R = Vxi*t = 933.6*33.57 = 31,340 m

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